Free math worksheets

- The radius is the distance from the x-axis to the graph. - Square the radius to obtain each disc's area. - Accumulate the constant disc area across the interval.

1. Every disc has radius \(2\), so its area is \(\pi(2)^2=4\pi\). 2. Therefore, \(V=\pi\int_0^5 2^2\,\mathrm{d}x=4\pi(5)=20\pi\).

\(20\pi\) cubic units

- Recall the disc-method formula for rotation about the x-axis. - Simplify the squared square root. - Find an antiderivative of the resulting linear function. - Include the factor \(\pi\).

1. By the disc method, \(V=\pi\int_1^5(\sqrt{3x+1})^2\,\text{d}x=\pi\int_1^5(3x+1)\,\text{d}x\). 2. \(V=\pi[\frac{3}{2}x^2+x]_1^5=\pi(42.5-2.5)=40\pi\).

\(40\pi\) cubic units

- Use the displayed vertical distance as the disc radius. - Square the entire reciprocal expression. - Rewrite the integrand with a negative exponent before applying the power rule.

1. The disc radius is \(\frac{2}{x}\), so \(V=\pi\int_1^4\left(\frac{2}{x}\right)^2\,\mathrm{d}x=4\pi\int_1^4x^{-2}\,\mathrm{d}x\). 2. Therefore, \(V=4\pi\left[-\frac{1}{x}\right]_1^4=3\pi\) cubic units.

\(3\pi\) cubic units

- Use the displayed vertical distance as the disc radius. - Squaring the radius removes the square root. - Use the given endpoints as the integration limits.

1. The disc radius is \(\sqrt{x}\), so \(V=\pi\int_1^4(\sqrt{x})^2\,\mathrm{d}x=\pi\int_1^4x\,\mathrm{d}x\). 2. Therefore, \(V=\pi\left[\frac{x^2}{2}\right]_1^4=\frac{15\pi}{2}\) cubic units.

\(\frac{15\pi}{2}\) cubic units

- Use the displayed vertical distance as the disc radius. - Squaring the radius removes the square root. - Integrate the resulting polynomial term by term.

1. The disc radius is \(\sqrt{x^2+5}\), so \(V=\pi\int_0^2(\sqrt{x^2+5})^2\,\mathrm{d}x=\pi\int_0^2(x^2+5)\,\mathrm{d}x\). 2. Therefore, \(V=\pi\left[\frac{x^3}{3}+5x\right]_0^2=\frac{38\pi}{3}\) cubic units.

\(\frac{38\pi}{3}\) cubic units

- Translate the stated average into an accumulated quantity over the interval. - Identify which squared measurement determines each disc's area. - Keep the factor common to every circular cross section until the end.

1. The average-value statement gives \(\frac15\int_0^5[R(y)]^2\,\mathrm{d}y=7\). 2. Therefore, \(\int_0^5[R(y)]^2\,\mathrm{d}y=35\). 3. The disc-method volume is \(V=\pi\int_0^5[R(y)]^2\,\mathrm{d}y=35\pi\).

\(35\pi\) cubic units

- Use the figure to identify the distance from the axis of rotation to the boundary. - A disc radius is measured from the axis to one edge, not across the full solid. - Replace the incorrect radius before evaluating the integral.

1. The displayed vertical distance \(6-x\) runs from the axis of rotation to the boundary, so it is already the disc radius, not a diameter. 2. The correct integral is \(V=\pi\int_0^6(6-x)^2\,\mathrm{d}x\). 3. Evaluating gives \(V=72\pi\) cubic units.

The radius was incorrectly halved. The correct volume is \(72\pi\) cubic units.

- Track how a vertical scale changes every disc radius. - Volume depends on the square of the radius scale.

1. The original volume is \(\pi\int_a^b[f(x)]^2\,\mathrm{d}x=12\pi\). 2. For \(g(x)=3f(x)\), the squared radius is \(9[f(x)]^2\). 3. Therefore, the new volume is \(9(12\pi)=108\pi\).

\(108\pi\) cubic units

- What solid is formed when a slanted line segment is rotated about an axis? - Recall the disc-method formula for rotation about the x-axis. - Square both the variable and the coefficient in \(f(x)\). - Substitute both interval endpoints carefully.

1. Rotating the line segment over an interval that does not include the origin creates a conical frustum. 2. By the disc method, \(V=\pi\int_k^{3k}\left(\frac{x}{2}\right)^2\,\text{d}x=\frac{\pi}{4}\int_k^{3k}x^2\,\text{d}x\). 3. \(V=\frac{\pi}{12}[(3k)^3-k^3]=\frac{\pi}{12}(26k^3)=\frac{13\pi}{6}k^3\).

The solid is a conical frustum with volume \(V=\frac{13\pi}{6}k^3\).

- Recall the disc-method formula for rotation about the x-axis. - What happens when \(\sqrt{rx}\) is squared? - Where is the radius largest on \([0,r]\)? - Use the cylinder-volume formula for the comparison.

1. By the disc method, \(V=\pi\int_0^r(\sqrt{rx})^2\,\text{d}x=\pi\int_0^r rx\,\text{d}x\). 2. \(V=\pi r[\frac{1}{2}x^2]_0^r=\frac{1}{2}\pi r^3\). 3. The maximum radius occurs at \(x=r\), where \(f(r)=r\). The comparison cylinder has radius \(r\) and height \(r\), so its volume is \(V_{\text{cyl}}=\pi r^3\). 4. Therefore, the solid of revolution has half the volume of the cylinder.

The solid has volume \(V=\frac{1}{2}\pi r^3\), which is half the cylinder's volume \(\pi r^3\).

- Identify the function that is squared in the volume integral. - What solid forms when a slanted line segment is rotated about an axis? - Use the power rule after treating \(x+2\) as a single expression. - The endpoint radii are the function values at the integration limits.

1. Comparing with \(V=\pi\int_a^b[f(x)]^2\,\text{d}x\) gives \(f(x)=x+2\). Rotating this line segment about the x-axis produces a right conical frustum. 2. \(V=\pi[\frac{1}{3}(x+2)^3]_0^3=\pi\left(\frac{125}{3}-\frac{8}{3}\right)=39\pi\). 3. The endpoint radii are \(f(0)=2\) and \(f(3)=5\).

1. \(f(x)=x+2\); a right conical frustum 2. \(39\pi\) 3. \(2\) and \(5\)

- First write the paraboloid's volume as a function of \(a\). - Simplify \((\sqrt{x})^2\). - When squaring \(\frac{1}{2}x\), square both the coefficient and the variable. - Simplify the ratio by canceling common factors.

1. \(V_{\text{paraboloid}}=\pi\int_0^a(\sqrt{x})^2\,\text{d}x=\pi\int_0^a x\,\text{d}x=\frac{1}{2}\pi a^2\). Setting this equal to \(32\pi\) gives \(a^2=64\), so \(a=8\). 2. \(V_{\text{cone}}=\pi\int_0^8\left(\frac{x}{2}\right)^2\,\text{d}x=\frac{\pi}{4}[\frac{x^3}{3}]_0^8=\frac{128\pi}{3}\). 3. \(32\pi:\frac{128\pi}{3}=3:4\).

1. \(a=8\) 2. \(V_{\text{cone}}=\frac{128\pi}{3}\) cubic units 3. \(3:4\)

- Write the volume of the liquid as a disc-method integral. - Recall the relationship between milliliters and cubic centimeters. - The liquid height is the upper limit of integration. - After integrating, solve the resulting equation for \(h\).

1. If the liquid reaches height \(h\), its volume is \(V(h)=\pi\int_0^h(f(x))^2\,\text{d}x\). 2. \(V(h)=\pi\int_0^h3x\,\text{d}x=\frac{3}{2}\pi h^2\). 3. Since \(150\,\text{mL}=150\,\text{cm}^3\), solve \(\frac{3}{2}\pi h^2=150\). 4. \(h^2=\frac{100}{\pi}\), so \(h=\sqrt{\frac{100}{\pi}}\approx5.64\,\text{cm}\).

The liquid is approximately \(5.64\,\text{cm}\) deep.

- Simplify the square of the exponential function before integrating. - Remember that \(e^0=1\) when evaluating the lower limit. - Use the natural logarithm to solve an exponential equation. - Check that the resulting height lies in \([0,8]\).

1. If the water reaches height \(h\), its volume is \(V(h)=\pi\int_0^h(e^{0.2x})^2\,\mathrm{d}x=\pi\int_0^h e^{0.4x}\,\mathrm{d}x\). 2. \(V(h)=\frac{\pi}{0.4}(e^{0.4h}-1)=2.5\pi(e^{0.4h}-1)\). 3. Since \(60\,\text{mL}=60\,\text{cm}^3\), solve \(2.5\pi(e^{0.4h}-1)=60\). 4. \(e^{0.4h}=1+\frac{24}{\pi}\), so \(h=\frac{1}{0.4}\ln\left(1+\frac{24}{\pi}\right)\approx5.39\,\text{cm}\). This value is within the modeled height \([0,8]\).

The water is approximately \(5.39\,\text{cm}\) deep.

- Use the displayed vertical distance as the radius of a representative disc. - Square the entire radius function before integrating. - Simplify the squared radical to obtain a polynomial.

1. A representative disc has radius \(f(x)=x\sqrt{6-x}\), so \(V=\pi\int_0^6[f(x)]^2\,\mathrm{d}x\). 2. \((x\sqrt{6-x})^2=x^2(6-x)=6x^2-x^3\). 3. Therefore, \(V=\pi\left[2x^3-\frac{1}{4}x^4\right]_0^6=108\pi\) cubic units.

\(108\pi\) cubic units

- Identify the left and right boundaries of the region. - Square the entire function before integrating. - Rewrite the reciprocal square using a negative exponent. - Apply the power rule with the linear expression \(x+1\).

1. The bounds are \(x=0\) and \(x=2\). 2. By the disc method, \(V=\pi\int_0^2\left(\frac{3}{x+1}\right)^2\,\mathrm{d}x=9\pi\int_0^2(x+1)^{-2}\,\mathrm{d}x\). 3. \(V=\pi[-\frac{9}{x+1}]_0^2=\pi(-3+9)=6\pi\).

\(6\pi\) cubic units

- Square both the coefficient and the exponential factor. - Integrate the exponential function with its linear exponent. - Check signs and use \(e^0=1\) when evaluating the limits. - Give an exact value before rounding.

1. By the disc method, \(V=\pi\int_0^4(2e^{-0.25x})^2\,\text{d}x=4\pi\int_0^4e^{-0.5x}\,\text{d}x\). 2. \(V=4\pi[-2e^{-0.5x}]_0^4=8\pi(1-e^{-2})\).

\(V=8\pi(1-e^{-2})\) cubic units, approximately \(21.73\) cubic units

- Use the upper semicircle as the radius function for the discs. - Squaring \(\sqrt{R^2-x^2}\) removes the radical. - Use symmetry or evaluate the antiderivative at \(-R\) and \(R\).

1. Rotating the upper semicircle about the x-axis produces discs with radius \(\sqrt{R^2-x^2}\). 2. Therefore, \(V=\pi\int_{-R}^R(R^2-x^2)\,\mathrm{d}x\). 3. Evaluating gives \(V=\pi\left[R^2x-\frac{x^3}{3}\right]_{-R}^R=\frac{4}{3}\pi R^3\).

\(V=\frac{4}{3}\pi R^3\)

- Rewrite the fourth root using a rational exponent. - Use the disc-method volume formula. - Treat both integrals as improper. - Compare the behavior of reciprocal powers as \(x\to\infty\) and as \(x\to0^+\).

1. \((f(x))^2=\frac{4}{x^{3/2}}=4x^{-3/2}\), and an antiderivative is \(-8x^{-1/2}\). 2. Over \([1,\infty)\), \(V_1=\pi\lim_{b\to\infty}[-8x^{-1/2}]_1^b=\pi(0+8)=8\pi\). This volume is finite. 3. Over \((0,1]\), \(V_2=\pi\lim_{a\to0^+}[-8x^{-1/2}]_a^1\). Since \(8a^{-1/2}\to\infty\), the volume diverges.

Over \([1,\infty)\), the volume is \(8\pi\) cubic units. Over \((0,1]\), the volume is infinite.

- Use the disc-method formula with the liquid height as the upper limit. - Compare the combined volume of the liquid and submerged marbles with the full capacity. - Use the sphere-volume formula for each marble. - Evaluate the antiderivative at \(h\) to remove the integral sign.

1. At height \(8\), the liquid volume is \(V(8)=\pi\int_0^8(0.2x+3)^2\,\mathrm{d}x\). 2. \(V(8)=\pi\left[\frac{x^3}{75}+0.6x^2+9x\right]_0^8=\frac{8792\pi}{75}\approx368.28\,\text{cm}^3\). 3. The full capacity is \(V(10)=\pi\left[\frac{x^3}{75}+0.6x^2+9x\right]_0^{10}=\frac{490\pi}{3}\approx513.13\,\text{cm}^3\). 4. The five marbles displace \(5\left(\frac{4}{3}\pi(1.2)^3\right)=\frac{288\pi}{25}\approx36.19\,\text{cm}^3\). 5. The liquid and marbles occupy approximately \(368.28+36.19=404.47\,\text{cm}^3\), which is less than the capacity, so the container does not overflow. 6. Replacing the upper limit by \(h\) gives \(\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)=300\).

a) Approximately \(368.28\,\text{cm}^3\) b) No. The liquid and marbles occupy approximately \(404.47\,\text{cm}^3\), less than the capacity of approximately \(513.13\,\text{cm}^3\). c) \(\pi\left(\frac{h^3}{75}+0.6h^2+9h\right)=300\)

- Determine the slope from the two endpoint radii. - Use the line as the disc radius function. - Treat \(r\) and \(h\) as constants while integrating with respect to \(x\).

1. The line passes through \((0,0)\) and \((h,r)\), so its slope is \(m=\frac{r}{h}\). Thus, \(f(x)=\frac{r}{h}x\). 2. By the disc method, \(V=\pi\int_0^h\left(\frac{r}{h}x\right)^2\,\mathrm{d}x\). 3. Therefore, \(V=\pi\frac{r^2}{h^2}\left[\frac{x^3}{3}\right]_0^h=\frac{1}{3}\pi r^2h\).

a) \(f(x)=\frac{r}{h}x\) b) \(V=\frac{1}{3}\pi r^2h\)

- Use the given radius and height as a point on the profile. - Solve the profile equation for radius \(x\) in terms of height \(y\). - A horizontal slice rotates into a disc whose radius is the displayed horizontal distance.

1. The point \((4,8)\) lies on the profile, so \(8=a(4)^3\). Therefore, \(a=\frac{1}{8}=0.125\). 2. Solve for the radius in terms of height: \(x=(8y)^{1/3}\), so \(x^2=4y^{2/3}\). 3. Rotating about the y-axis gives discs with radius \(x(y)\), so \(V=\pi\int_0^8x(y)^2\,\mathrm{d}y=4\pi\int_0^8y^{2/3}\,\mathrm{d}y\). 4. Therefore, \(V=\frac{384\pi}{5}\approx241.27\,\text{cm}^3\).

a) \(a=\frac{1}{8}=0.125\) b) \(\frac{384\pi}{5}\,\text{cm}^3\approx241.27\,\text{cm}^3\)

- Substitute the radius function into the disc-method formula. - Simplify the squared radical before integrating. - Use the logarithmic antiderivative for a reciprocal linear expression. - Apply the exponential function to undo the logarithm.

1. By the disc method, \(V(b)=\pi\int_0^b\left(\frac{3}{\sqrt{2x+1}}\right)^2\,\mathrm{d}x=9\pi\int_0^b\frac{1}{2x+1}\,\mathrm{d}x\). 2. Therefore, \(V(b)=\frac{9\pi}{2}[\ln(2x+1)]_0^b=\frac{9\pi}{2}\ln(2b+1)\). 3. Set the formula equal to \(9\pi\): \(\frac{9\pi}{2}\ln(2b+1)=9\pi\). 4. Then \(\ln(2b+1)=2\), so \(2b+1=e^2\) and \(b=\frac{e^2-1}{2}\approx3.19\).

a) \(V(b)=\frac{9\pi}{2}\ln(2b+1)\) b) \(b=\frac{e^2-1}{2}\approx3.19\)

- Use the zeros of the function to determine the interval. - Apply the disc-method formula. - Expand the square of the function before integrating. - Be careful with signs when evaluating at the lower limit.

1. The zeros are \(x=-1\) and \(x=2\), so these are the limits of integration. 2. By the disc method, \(V=\pi\int_{-1}^{2}(f(x))^2\,\mathrm{d}x\). 3. \((f(x))^2=(x+1)^2(2-x)=-x^3+3x+2\). 4. Therefore, \(V=\pi\left[-\frac{x^4}{4}+\frac{3x^2}{2}+2x\right]_{-1}^{2}\). 5. The definite integral is \(6-(-\frac{3}{4})=\frac{27}{4}\), so \(V=\frac{27\pi}{4}\approx21.21\) cubic units.

\(\frac{27\pi}{4}\) cubic units, or approximately \(21.21\) cubic units

- Squaring the radius function removes the square root. - Use integration by parts for the product \(x\sin x\). - Recall the sine and cosine values at \(0\) and \(\pi\). - Include the factor \(\pi\) from the disc-method formula.

1. By the disc method, \(V=\pi\int_0^\pi(f(x))^2\,\mathrm{d}x=\pi\int_0^\pi x\sin x\,\mathrm{d}x\). 2. Use integration by parts with \(u=x\) and \(\mathrm{d}v=\sin x\,\mathrm{d}x\): \(\int x\sin x\,\mathrm{d}x=-x\cos x+\sin x\). 3. \(\int_0^\pi x\sin x\,\mathrm{d}x=[-x\cos x+\sin x]_0^\pi=\pi\). 4. Therefore, \(V=\pi(\pi)=\pi^2\) cubic units.

\(\pi^2\) cubic units

- Use the disc-method formula. - Simplify the squared square root before integrating. - Use mass equals density times volume. - Keep the units consistent.

1. By the disc method, \(V=\pi\int_0^5(2\sqrt{x})^2\,\mathrm{d}x=\pi\int_0^5 4x\,\mathrm{d}x\). 2. \(V=\pi[2x^2]_0^5=50\pi\approx157.08\,\text{cm}^3\). 3. Using \(m=\rho V\), \(m=0.8(50\pi)=40\pi\approx125.66\,\text{g}\).

1. \(50\pi\,\text{cm}^3\approx157.08\,\text{cm}^3\) 2. \(40\pi\,\text{g}\approx125.66\,\text{g}\)

- Use the disc-method formula. - Square both factors in the radius function. - Use integration by parts for \(xe^{2x}\). - Include both terms of the antiderivative when evaluating the lower limit.

1. By the disc method, \(V=\pi\int_0^1(f(x))^2\,\mathrm{d}x=\pi\int_0^1xe^{2x}\,\mathrm{d}x\). 2. Integration by parts gives \(\int xe^{2x}\,\mathrm{d}x=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}\). 3. Therefore, \(V=\pi\left[\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}\right]_0^1\). 4. Evaluating the limits gives \(V=\pi\left(\frac{e^2}{4}+\frac{1}{4}\right)=\frac{\pi(e^2+1)}{4}\).

\(\frac{\pi(e^2+1)}{4}\) cubic units

- Use the intercepts to identify the left and right ends of the region. - The two curved boundaries are reflections across the y-axis. - Use one half of the region and double the disc-method integral.

1. The x-intercepts are \(x=\ln2\) for \(f\) and \(x=-\ln2\) for \(g\). The graphs meet at \((0,1)\). 2. Since \(f(x)=g(-x)\), the region is symmetric about the y-axis. Therefore, \(V=2\pi\int_0^{\ln2}(2-e^x)^2\,\mathrm{d}x\). 3. Expanding gives \((2-e^x)^2=4-4e^x+e^{2x}\), with antiderivative \(4x-4e^x+\frac{1}{2}e^{2x}\). 4. Evaluating gives \(V=\pi(8\ln2-5)\approx1.71\) cubic units.

\(\pi(8\ln2-5)\) cubic units, or approximately \(1.71\) cubic units

- Find where the sine and cosine graphs intersect. - Identify the upper boundary on each part of the interval. - Use the symmetry of sine and cosine on this interval. - Apply a power-reduction identity to integrate the squared trigonometric function.

1. The graphs intersect where \(\sin x=\cos x\), so \(x=\frac{\pi}{4}\). 2. The upper boundary is \(\sin x\) on \([0,\frac{\pi}{4}]\) and \(\cos x\) on \([\frac{\pi}{4},\frac{\pi}{2}]\). 3. By symmetry, \(V=2\pi\int_0^{\pi/4}\sin^2x\,\mathrm{d}x\). 4. Using \(\sin^2x=\frac{1}{2}(1-\cos2x)\), \(V=\pi\left[x-\frac{1}{2}\sin2x\right]_0^{\pi/4}\). 5. Therefore, \(V=\frac{\pi^2}{4}-\frac{\pi}{2}\approx0.90\) cubic units.

\(\frac{\pi^2}{4}-\frac{\pi}{2}\) cubic units, or approximately \(0.90\) cubic units

- Differentiate the radius function to locate its maximum on \([0,4]\). - Check the derivative sign on both sides of the critical point. - Square the entire radius function before integrating.

1. For \(x>0\), \(g'(x)=\frac{4-3x}{2\sqrt{x}}\). 2. The derivative changes from positive to negative at \(x=\frac{4}{3}\), so the maximum radius is \(g\left(\frac{4}{3}\right)=\frac{16\sqrt3}{9}\). 3. By the disc method, \(V=\pi\int_0^4[(4-x)\sqrt{x}]^2\,\mathrm{d}x=\pi\int_0^4(16x-8x^2+x^3)\,\mathrm{d}x\). 4. Therefore, \(V=\frac{64\pi}{3}\approx67.02\) cubic units.

a) The maximum radius is \(\frac{16\sqrt3}{9}\), occurring at \(x=\frac{4}{3}\). b) \(\frac{64\pi}{3}\) cubic units, or approximately \(67.02\) cubic units

- Reflecting across the x-axis changes the sign of each function value. - The narrowest point corresponds to the minimum radius. - Use the disc-method formula for capacity and a cylinder formula for the attached base.

1. Reflection across the x-axis gives \(g(x)=-f(x)=-0.1x^2+x-5\). 2. Since \(f'(x)=0.2x-1\), the minimum occurs at \(x=5\). Because \(f(5)=2.5\), the narrowest diameter is \(5\,\text{cm}\). 3. Since \(f(8)=3.4\), the opening diameter is \(6.8\,\text{cm}\). 4. By the disc method, \(V=\pi\int_0^8(0.1x^2-x+5)^2\,\mathrm{d}x=\frac{30{,}776\pi}{375}\approx257.8\,\text{cm}^3\). 5. The base has radius \(5\,\text{cm}\) and thickness \(0.8\,\text{cm}\), so its volume is \(20\pi\approx62.8\,\text{cm}^3\).

a) \(g(x)=-0.1x^2+x-5\) b) \(5\,\text{cm}\) c) \(6.8\,\text{cm}\) d) \(257.8\,\text{cm}^3\) e) \(62.8\,\text{cm}^3\)

- Use the x-intercepts as the integration limits. - Square the displayed radius function before integrating. - Use even symmetry to integrate over half the interval and double.

1. The x-intercepts are \(-2\) and \(2\), so \(V=\pi\int_{-2}^{2}(4-x^2)^2\,\mathrm{d}x\). 2. Expand the integrand: \((4-x^2)^2=16-8x^2+x^4\). 3. Using symmetry, \(V=2\pi\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^2=\frac{512\pi}{15}\) cubic units.

\(\frac{512\pi}{15}\) cubic units

- Use the displayed sine value as the disc radius. - Apply a power-reduction identity to \(\sin^2x\). - Keep the factor \(\pi\) from the disc-area formula.

1. By the disc method, \(V=\pi\int_0^\pi\sin^2x\,\mathrm{d}x\). 2. Use \(\sin^2x=\frac{1}{2}(1-\cos2x)\). 3. Therefore, \(V=\pi\left[\frac{x}{2}-\frac{\sin2x}{4}\right]_0^\pi=\frac{\pi^2}{2}\) cubic units.

\(\frac{\pi^2}{2}\) cubic units

- Measure each radius from the dashed horizontal axis of rotation. - The curve lies on opposite sides of the axis over the two halves of the interval. - Squaring the radius allows one integral over the full interval.

1. The axis of rotation is a boundary of the region, so each perpendicular cross section is a disc. 2. The radius is the distance \(|\cos x|\), and its square is \(\cos^2x\). 3. Therefore, \(V=\pi\int_0^\pi\cos^2x\,\mathrm{d}x=\frac{\pi^2}{2}\) cubic units.

\(\frac{\pi^2}{2}\) cubic units

- Use slices perpendicular to the vertical axis of rotation. - Rewrite \(y=e^x\) as \(x=\ln y\). - The horizontal distance from the y-axis to the curve is the disc radius.

1. Rewrite the curve as \(x=\ln y\) for \(1\le y\le e\). 2. A horizontal slice rotates into a disc with radius \(R(y)=\ln y\). 3. Therefore, \(V=\pi\int_1^e(\ln y)^2\,\mathrm{d}y=\pi(e-2)\) cubic units.

\(\pi(e-2)\) cubic units

- The scale factor in the radius is squared in the volume. - Use the positivity condition after solving the equation.

1. The disc radius is \(kx\), so \(V=\pi\int_0^3k^2x^2\,\mathrm{d}x=9\pi k^2\). 2. Set \(9\pi k^2=81\pi\). 3. Since \(k>0\), \(k=3\).

\(k=3\)

- Use horizontal slices because the axis of rotation is vertical. - The horizontal distance from the y-axis to the curve is the disc radius. - Square the entire radius expression before integrating.

1. A horizontal slice forms a disc with radius \(R(y)=y(4-y)\). 2. Therefore, \(V=\pi\int_0^4y^2(4-y)^2\,\mathrm{d}y\). 3. Expanding and integrating gives \(\int_0^4(16y^2-8y^3+y^4)\,\mathrm{d}y=\frac{512}{15}\). 4. Thus, \(V=\frac{512\pi}{15}\) cubic units.

\(\frac{512\pi}{15}\) cubic units

- Recall the disc-method formula for a solid formed by rotation about the x-axis. - Rewrite the improper integral using a variable upper limit. - When does \(x^n\) approach zero as \(x\to\infty\)? - Check separately the case in which the antiderivative is logarithmic.

1. Using the disc method, \(V=\pi\int_1^\infty(f_k(x))^2\,\mathrm{d}x=\pi\int_1^\infty x^{-2k}\,\mathrm{d}x\). 2. When \(k\ne\frac{1}{2}\), an antiderivative is \(\frac{x^{1-2k}}{1-2k}\). The improper integral converges only when \(1-2k<0\), or \(k>\frac{1}{2}\). 3. When \(k=\frac{1}{2}\), the integrand is \(x^{-1}\), and the integral diverges logarithmically. 4. For \(k>\frac{1}{2}\), \(\int_1^\infty x^{-2k}\,\mathrm{d}x=\frac{1}{2k-1}\). Therefore, \(V(k)=\frac{\pi}{2k-1}\).

The volume is finite for \(k>\frac{1}{2}\), and \(V(k)=\frac{\pi}{2k-1}\).

- Treat each integral as improper because the function is undefined at the left endpoint. - Find antiderivatives of \(x^{-1/2}\) and \(\frac{1}{x}\). - Compare the behavior of \(\sqrt{x}\) and \(\ln x\) as \(x\to0^+\).

1. The area is the improper integral \(A=\int_0^1x^{-1/2}\,\mathrm{d}x=\lim_{a\to0^+}[2\sqrt{x}]_a^1=2\). 2. The volume is \(V=\pi\int_0^1(g(x))^2\,\mathrm{d}x=\pi\int_0^1\frac{1}{x}\,\mathrm{d}x\). 3. \(V=\pi\lim_{a\to0^+}[\ln x]_a^1=\pi\lim_{a\to0^+}(0-\ln a)=\infty\). 4. Thus, the region has finite area, but its solid of revolution has infinite volume.

The area is \(2\) square units. The volume of the solid of revolution diverges and is infinite.

- Express the upper integration limit using the bottom location and the water depth. - A depth of \(h\) places the water surface at \(x=-6+h\). - A fully submerged object has the same displaced volume before and after more water is added. - Set the difference between the two container volumes equal to the added volume.

1. For a water depth \(h\), the volume is \(V(h)=\pi\int_{-6}^{-6+h}(f(x))^2\,\mathrm{d}x=12\int_{-6}^{-6+h}\cos\left(\frac{\pi x}{12}\right)\,\mathrm{d}x\). 2. Therefore, \(V(h)=\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-6)}{12}\right)+1\right)\). 3. At \(h=3\), \(V(3)=\frac{144}{\pi}\left(1-\frac{\sqrt{2}}{2}\right)\approx13.43\,\text{in.}^3\). 4. Because the object is completely submerged both before and after the added water, its displaced volume cancels. Thus, \(V(h+1)-V(h)=4\). 5. Substitution gives \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\).

a) Approximately \(13.43\,\text{in.}^3\) b) \(\frac{144}{\pi}\left(\sin\left(\frac{\pi(h-5)}{12}\right)-\sin\left(\frac{\pi(h-6)}{12}\right)\right)=4\)

- First find the equation of the line through the two given endpoint radii. - Use the disc-method formula. - Expand \((ax+b)^2\) before integrating. - Simplify the resulting polynomial in \(R\) and \(r\).

1. The line through \((0,r)\) and \((h,R)\) is \(f(x)=\frac{R-r}{h}x+r\). 2. By the disc method, \(V=\pi\int_0^h\left(\frac{R-r}{h}x+r\right)^2\,\mathrm{d}x\). 3. Expanding and integrating gives \(V=\pi\left[\frac{(R-r)^2}{3h^2}x^3+\frac{r(R-r)}{h}x^2+r^2x\right]_0^h\). 4. Thus, \(V=\pi h\left(\frac{(R-r)^2}{3}+r(R-r)+r^2\right)\). 5. Simplifying the expression in parentheses gives \(\frac{R^2+Rr+r^2}{3}\), so \(V=\frac{1}{3}\pi h(R^2+Rr+r^2)\).

\(V=\frac{1}{3}\pi h(R^2+Rr+r^2)\)

- Identify the antiderivative needed for each integral. - How does the natural logarithm behave as its input increases without bound? - Compare the convergence of reciprocal first and second powers. - An unbounded region can still generate a solid with finite volume.

1. The area is \(A=\int_0^\infty\frac{1}{x+2}\,\mathrm{d}x=\lim_{b\to\infty}[\ln(x+2)]_0^b\). 2. Since \(\ln(b+2)-\ln2\to\infty\), the area diverges. 3. The volume is \(V=\pi\int_0^\infty\frac{1}{(x+2)^2}\,\mathrm{d}x\). 4. \(V=\pi\lim_{b\to\infty}[-\frac{1}{x+2}]_0^b=\pi\left(0+\frac{1}{2}\right)=\frac{\pi}{2}\).

a) The area is infinite. b) The volume is \(\frac{\pi}{2}\) cubic units.

- Relate one model unit to the corresponding actual length. - Use the disc-method formula for rotation about the x-axis. - Remember that scaling lengths by \(10\) scales volumes by \(10^3\). - Use mass equals density times volume. - Subtract the volume of the cylindrical recess from the original volume.

1. The height is \(4-(-4)=8\) model units, so the actual height is \(8\cdot10=80\,\text{cm}\). 2. The seat radius is \(f(4)=\sqrt{0.2(4)^2+4}=\sqrt{7.2}\) model units. The actual diameter is \(2\sqrt{7.2}\cdot10\approx53.67\,\text{cm}\). 3. By the disc method, \(V=\pi\int_{-4}^{4}(f(x))^2\,\mathrm{d}x=\pi\int_{-4}^{4}(0.2x^2+4)\,\mathrm{d}x\). 4. \(V=\pi\left[\frac{x^3}{15}+4x\right]_{-4}^{4}=\frac{608\pi}{15}\) model cubic units. 5. Since one model cubic unit is \(1000\,\text{cm}^3=1\,\text{L}\), the actual volume is \(\frac{608\pi}{15}\approx127.34\,\text{L}\). 6. The mass is \(0.095\cdot127{,}339.22\approx12{,}097.23\,\text{g}\approx12.10\,\text{kg}\). 7. The recess has model volume \(\pi(1)^2(2)=2\pi\). The remaining volume is \(\frac{608\pi}{15}-2\pi=\frac{578\pi}{15}\approx121.06\,\text{L}\).

a) Height: \(80\,\text{cm}\); seat diameter: approximately \(53.67\,\text{cm}\) b) Approximately \(127.34\,\text{L}\) c) Approximately \(12.10\,\text{kg}\) d) Approximately \(121.06\,\text{L}\)

- Use the displayed radius function in the disc-method formula. - Express the bowl volume below a variable height \(h\). - Check that the cylinder fits by comparing its radius with the bowl radius at the bottom. - The submerged part of the cylinder displaces an equal volume of water.

1. The full capacity is \(V_{\max}=\pi\int_0^{12}(\sqrt{x+4})^2\,\mathrm{d}x=\pi\int_0^{12}(x+4)\,\mathrm{d}x\). 2. Therefore, \(V_{\max}=\pi\left[\frac{x^2}{2}+4x\right]_0^{12}=120\pi\approx376.99\,\text{cm}^3\). 3. For a water depth \(h\), \(\pi\left(\frac{h^2}{2}+4h\right)=150\). The positive solution is \(h=-4+\sqrt{16+\frac{300}{\pi}}\approx6.56\,\text{cm}\). 4. The cylinder fits because its radius \(1\) is less than the bowl radius \(g(0)=2\) at the bottom. When submerged to depth \(h_{\text{new}}\), it occupies \(\pi h_{\text{new}}\) cubic centimeters below the water surface. 5. Therefore, the bowl volume below the new surface equals the water volume plus the submerged cylinder volume: \(\pi\left(\frac{h_{\text{new}}^2}{2}+4h_{\text{new}}\right)=150+\pi h_{\text{new}}\).

a) \(120\pi\,\text{cm}^3\approx376.99\,\text{cm}^3\) b) Approximately \(6.56\,\text{cm}\) c) The cylinder reduces the space available to the water, so the surface must rise. An equation for the new depth is \(\pi\left(\frac{h_{\text{new}}^2}{2}+4h_{\text{new}}\right)=150+\pi h_{\text{new}}\).

- Reflection across the x-axis changes the sign of the radius function. - The maximum diameter is twice the maximum radius shown by the profile. - Use the disc-method formula, then match the box dimensions to the axial length and maximum diameter.

1. Reflection across the x-axis gives \(g(x)=-0.1(x+2)\sqrt{25-x}\). 2. Differentiating shows that the maximum radius occurs at \(x=16\). Since \(f(16)=5.4\), the maximum diameter is \(10.8\,\text{cm}\). 3. By the disc method, \(V=\pi\int_0^{25}[0.1(x+2)\sqrt{25-x}]^2\,\mathrm{d}x=\frac{7075\pi}{16}\approx1389.17\,\text{cm}^3\). 4. The axial length is \(25\,\text{cm}\), and the maximum width and height are both \(10.8\,\text{cm}\).

a) \(g(x)=-0.1(x+2)\sqrt{25-x}\) b) \(10.8\,\text{cm}\) c) \(\frac{7075\pi}{16}\,\text{cm}^3\approx1389.17\,\text{cm}^3\) d) \(25\,\text{cm}\times10.8\,\text{cm}\times10.8\,\text{cm}\)