Free math worksheets

- Use the accumulated squared radii directly. - Subtract the inner contribution from the outer contribution before multiplying by the circular factor.

1. Washer volume is \(V=\pi\int_a^b([R(x)]^2-[r(x)]^2)\,\mathrm{d}x\). 2. By linearity, \(V=\pi(30-12)=18\pi\).

\(18\pi\) cubic units

- The constant horizontal line gives the outer radius. - The square-root curve gives the inner radius. - Subtract the squared radii before integrating.

1. The curves meet at \(x=16\). On \([0,16]\), the outer radius is \(4\) and the inner radius is \(\sqrt{x}\). 2. Thus, \(V=\pi\int_0^{16}[4^2-(\sqrt{x})^2]\,\mathrm{d}x=\pi\int_0^{16}(16-x)\,\mathrm{d}x\). 3. Therefore, \(V=128\pi\approx402.12\) cubic units.

\(128\pi\) cubic units, or approximately \(402.12\) cubic units

- Decide which displayed function is farther from the x-axis. - Use the difference of the squared radii. - Rewrite the reciprocal square using a negative exponent.

1. On \([1,2]\), the outer radius is \(\frac{1}{x}\) and the inner radius is \(0.5\). 2. Therefore, \(V=\pi\int_1^2\left(\frac{1}{x^2}-0.25\right)\,\mathrm{d}x=\frac{\pi}{4}\approx0.785\) cubic units.

\(\frac{\pi}{4}\) cubic units, or approximately \(0.785\) cubic units

- Use the figure to identify which boundary is farther from the x-axis. - Washer area must be nonnegative on every slice. - Subtract the squared inner radius from the squared outer radius.

1. On \([0,3]\), \(4-x\ge1\), so the outer radius is \(4-x\) and the inner radius is \(1\). 2. The correct integral is \(V=\pi\int_0^3[(4-x)^2-1]\,\mathrm{d}x\). 3. Evaluating gives \(V=18\pi\) cubic units.

\(18\pi\) cubic units

- Compare distances to each axis rather than comparing the absolute y-coordinates. - Shift the region and the axis together. - If every cross-sectional area is unchanged, the accumulated volume is unchanged.

1. In the original solid, the outer and inner radii are \(f(x)\) and \(g(x)\). 2. In the shifted solid, the distances to \(y=5\) are again \(f(x)\) and \(g(x)\). 3. The washer areas are identical at every \(x\), with the same limits. 4. Therefore, the new volume is \(V\).

The new volume is \(V\).

- Identify the outer and inner radii from the axis of rotation. - Subtract the circular areas, not the radii, to form a washer. - Evaluate the incorrect thickness-squared expression separately for comparison.

1. The cross sections are washers with outer radius \(R(x)=\sqrt{x}+1\) and inner radius \(r(x)=1\). 2. Thus, \(V=\pi\int_0^4[(\sqrt{x}+1)^2-1^2]\,\mathrm{d}x=\frac{56\pi}{3}\) cubic units. 3. The comparison integral is \(V_{\text{incorrect}}=\pi\int_0^4(\sqrt{x})^2\,\mathrm{d}x=8\pi\) cubic units. 4. The values differ because washer area is \(\pi(R^2-r^2)\), not \(\pi(R-r)^2\).

a) \(V=\frac{56\pi}{3}\) cubic units b) \(V_{\text{incorrect}}=8\pi\) cubic units. It is smaller because \(R^2-r^2\ne(R-r)^2\).

- Determine which function gives the outer radius on the interval. - Use the washer-method formula. - Simplify the integrand before integrating. - Expand the squared binomial carefully.

1. On \([1,3]\), \(f(x)\ge g(x)\ge0\), so the washer method gives \(V=\pi\int_1^3(f(x)^2-g(x)^2)\,\mathrm{d}x\). 2. \(V=\pi\int_1^3((x+1)^2-(\sqrt{2x+2})^2)\,\mathrm{d}x=\pi\int_1^3(x^2-1)\,\mathrm{d}x\). 3. \(V=\pi[\frac{x^3}{3}-x]_1^3=\pi\left(6+\frac{2}{3}\right)=\frac{20\pi}{3}\).

\(V=\frac{20\pi}{3}\) cubic units

- Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers with negative exponents before integrating. - Multiply the integral by \(\pi\).

1. On \([1,2]\), \(f(x)\ge g(x)\ge0\), so \(f\) is the outer radius and \(g\) is the inner radius. 2. \(V=\pi\int_1^2\left(\frac{1}{x^2}-\frac{1}{x^4}\right)\,\mathrm{d}x=\pi\int_1^2(x^{-2}-x^{-4})\,\mathrm{d}x\). 3. \(V=\pi[-x^{-1}+\frac{1}{3}x^{-3}]_1^2=\pi\left(-\frac{11}{24}+\frac{2}{3}\right)=\frac{5\pi}{24}\).

\(V=\frac{5\pi}{24}\) cubic units

- Use the disc-method formula for the exterior volume. - Square the interior-radius function before integrating. - Subtract the interior volume from the exterior volume. - The radius at a given \(x\) is the function value there.

1. The exterior volume is \(V_f=\pi\int_0^{10}(\sqrt{x+2})^2\,\mathrm{d}x=\pi\int_0^{10}(x+2)\,\mathrm{d}x=70\pi\,\text{cm}^3\). 2. The interior volume is \(V_g=\pi\int_0^{10}(0.2x+1)^2\,\mathrm{d}x=\frac{130\pi}{3}\,\text{cm}^3\). 3. The glass volume is \(V_f-V_g=70\pi-\frac{130\pi}{3}=\frac{80\pi}{3}\,\text{cm}^3\). 4. Solve \(\sqrt{x+2}=3\). Squaring gives \(x+2=9\), so \(x=7\,\text{cm}\).

a) \(70\pi\,\text{cm}^3\) b) \(\frac{130\pi}{3}\,\text{cm}^3\) c) \(\frac{80\pi}{3}\,\text{cm}^3\) d) \(x=7\,\text{cm}\)

- First find the container volume up to a variable water depth. - A solid object reduces the space available for water. - Determine whether the cylinder is partially or fully submerged at each depth. - Subtract the displaced cylinder volume from the container volume.

1. The container volume up to height \(h\) is \(V(h)=\pi\int_0^h(f(x))^2\,\mathrm{d}x=\pi\int_0^h(2x+4)\,\mathrm{d}x=\pi(h^2+4h)\). 2. At \(h=2\), the container volume is \(12\pi\,\text{in.}^3\). The cylinder occupies \(\pi(1)^2(2)=2\pi\,\text{in.}^3\), so the water volume is \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\). 3. At \(h=5\), the container volume is \(45\pi\,\text{in.}^3\). The entire cylinder remains submerged and displaces \(2\pi\,\text{in.}^3\), so the water volume is \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\).

a) \(10\pi\,\text{in.}^3\approx31.42\,\text{in.}^3\) b) \(43\pi\,\text{in.}^3\approx135.09\,\text{in.}^3\)

- Find the intersection points first. - Determine which function gives the outer radius. - Use the washer-method formula. - Expand the squared expressions carefully.

1. The curves intersect where \(x^2=4x-x^2\), giving \(x=0\) and \(x=2\). 2. On \([0,2]\), \(g(x)\ge f(x)\ge0\), so the washer method gives \(V=\pi\int_0^2(g(x)^2-f(x)^2)\,\mathrm{d}x\). 3. \((4x-x^2)^2-(x^2)^2=16x^2-8x^3\). 4. \(V=\pi[\frac{16}{3}x^3-2x^4]_0^2=\frac{32\pi}{3}\).

\(V=\frac{32\pi}{3}\) cubic units

- Find the intersection points to determine the bounds. - Use the difference of the squares of the outer and inner radii. - Rewrite reciprocal powers using negative exponents. - Check that the resulting volume is positive.

1. The curves intersect where \(\frac{2}{x}=3-x\), which gives \(x^2-3x+2=0\). Thus, the bounds are \(x=1\) and \(x=2\). 2. On \([1,2]\), \(g(x)\ge f(x)>0\), so \(V=\pi\int_1^2\left((3-x)^2-\left(\frac{2}{x}\right)^2\right)\,\mathrm{d}x\). 3. \(V=\pi\int_1^2(x^2-6x+9-4x^{-2})\,\mathrm{d}x\). 4. \(V=\pi[\frac{x^3}{3}-3x^2+9x+\frac{4}{x}]_1^2=\pi\left(\frac{32}{3}-\frac{31}{3}\right)=\frac{\pi}{3}\).

\(V=\frac{\pi}{3}\) cubic units

- Use the full radius profile for the pavilion volume. - For the open core, subtract the square of the inner radius from the square of the outer radius. - The floor slab is a circle whose radius is the profile value at \(x=0\).

1. The pavilion volume is \(V=\pi\int_0^5(4-0.1x^2)^2\,\mathrm{d}x=\frac{635\pi}{12}\approx166.24\,\text{m}^3\). 2. For \(0\le x\le3\), washers have outer radius \(g(x)\) and inner radius \(2\). Thus, \(V_{\text{between}}=\pi\int_0^3[(4-0.1x^2)^2-2^2]\,\mathrm{d}x\). 3. Evaluating gives \(V_{\text{between}}=\frac{14{,}643\pi}{500}\approx92.00\,\text{m}^3\). 4. The floor slab has area \(\pi[g(0)]^2=16\pi\,\text{m}^2\). The cost is \(45(16\pi)=720\pi\approx\$2261.95\).

a) \(\frac{635\pi}{12}\,\text{m}^3\approx166.24\,\text{m}^3\) b) \(\frac{14{,}643\pi}{500}\,\text{m}^3\approx92.00\,\text{m}^3\) c) \(720\pi\approx\$2261.95\)

- View the bowl as an outer solid with an inner cavity removed. - Use the different starting points of the two profiles when setting the bounds. - The capacity is the volume generated by the inner radius.

1. The outer volume is \(V_{\text{outer}}=\pi\int_0^{16}(0.5x+4)\,\mathrm{d}x=128\pi\,\text{cm}^3\). 2. The capacity is the inner volume: \(V_{\text{inner}}=\pi\int_4^{16}(0.8x-3.2)\,\mathrm{d}x=57.6\pi\approx180.96\,\text{cm}^3\). 3. Therefore, the glass volume is \(V_{\text{outer}}-V_{\text{inner}}=70.4\pi\approx221.17\,\text{cm}^3\).

The glass volume is \(70.4\pi\,\text{cm}^3\approx221.17\,\text{cm}^3\). The capacity is \(57.6\pi\,\text{cm}^3\approx180.96\,\text{cm}^3\).

- Subtract the inner cross-sectional area from the outer cross-sectional area. - Apply exponent rules when squaring each radius function. - Use the antiderivative of \(e^{kx}\). - The two integrals can be combined because they use the same interval.

1. Use washers: \(V=\pi\int_0^{10}(f(x)^2-g(x)^2)\,\mathrm{d}x\). 2. \(f(x)^2-g(x)^2=4e^{0.2x}-2.25e^{0.2x}=1.75e^{0.2x}\). 3. Therefore, \(V=1.75\pi\int_0^{10}e^{0.2x}\,\mathrm{d}x=1.75\pi[5e^{0.2x}]_0^{10}\). 4. Thus, \(V=8.75\pi(e^2-1)\approx175.63\,\text{cm}^3\).

\(8.75\pi(e^2-1)\,\text{cm}^3\approx175.63\,\text{cm}^3\)

- Use the intersections as the integration limits. - The sine curve is farther from the x-axis throughout the interior of the interval. - Apply a power-reduction identity to the squared sine term.

1. The curves intersect at \(x=0\) and \(x=\frac{\pi}{2}\). 2. On this interval, \(\sin x\) is the outer radius. Thus, \(V=\pi\int_0^{\pi/2}\left[\sin^2x-\left(\frac{2x}{\pi}\right)^2\right]\,\mathrm{d}x\). 3. The two component integrals are \(\int_0^{\pi/2}\sin^2x\,\mathrm{d}x=\frac{\pi}{4}\) and \(\int_0^{\pi/2}\frac{4x^2}{\pi^2}\,\mathrm{d}x=\frac{\pi}{6}\). 4. Therefore, \(V=\frac{\pi^2}{12}\approx0.822\) cubic units.

\(\frac{\pi^2}{12}\) cubic units, or approximately \(0.822\) cubic units

- Solve for the intersection points and check any solutions introduced by squaring. - Identify which function gives the outer radius. - Use the washer-method formula. - Subtract the squares of the functions, not the square of their difference.

1. Set the functions equal: \(\sqrt{x+4}=\frac{1}{2}x+2\). Squaring and checking the solutions gives intersections at \(x=-4\) and \(x=0\). 2. On \([-4,0]\), \(f(x)\ge g(x)\), and both functions are nonnegative. Thus, \(V=\pi\int_{-4}^{0}\left((\sqrt{x+4})^2-\left(\frac{x}{2}+2\right)^2\right)\,\mathrm{d}x\). 3. The integrand simplifies to \(-\frac{x^2}{4}-x\). 4. \(V=\pi\left[-\frac{x^3}{12}-\frac{x^2}{2}\right]_{-4}^{0}=\pi\left(\frac{8}{3}\right)=\frac{8\pi}{3}\approx8.38\) cubic units.

\(\frac{8\pi}{3}\) cubic units, or approximately \(8.38\) cubic units

- Find the intersection points to determine the interval. - Identify the outer and inner radii. - Square each function separately before subtracting. - A sketch can help confirm which graph is farther from the x-axis.

1. Solve \(x^2-4x+5=-x+5\). This gives \(x(x-3)=0\), so the curves intersect at \(x=0\) and \(x=3\). 2. On \([0,3]\), \(g(x)\ge f(x)>0\), so \(g\) is the outer radius and \(f\) is the inner radius. 3. \(V=\pi\int_0^3(g(x)^2-f(x)^2)\,\mathrm{d}x\). 4. The difference is \((-x+5)^2-(x^2-4x+5)^2=-x^4+8x^3-25x^2+30x\). 5. \(V=\pi\left[-\frac{x^5}{5}+2x^4-\frac{25x^3}{3}+15x^2\right]_0^3=\frac{117\pi}{5}\approx73.51\) cubic units.

\(\frac{117\pi}{5}\) cubic units, or approximately \(73.51\) cubic units

- Find the intersection points first. - For area, integrate the upper function minus the lower function. - For volume, identify the outer and inner radii. - Square each radius function before subtracting.

1. Solving \(\frac{1}{2}x^2=x+4\) gives intersection points \(x=-2\) and \(x=4\). 2. On \([-2,4]\), \(g(x)\ge f(x)\). The area is \(A=\int_{-2}^{4}\left(x+4-\frac{x^2}{2}\right)\,\mathrm{d}x=18\) square units. 3. Both functions are nonnegative on the interval, so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 4. \(V=\pi\int_{-2}^{4}\left((x+4)^2-\left(\frac{x^2}{2}\right)^2\right)\,\mathrm{d}x\). 5. \(V=\pi\left[\frac{x^3}{3}+4x^2+16x-\frac{x^5}{20}\right]_{-2}^{4}=\frac{576\pi}{5}\approx361.91\) cubic units.

a) \(18\) square units b) \(\frac{576\pi}{5}\) cubic units, or approximately \(361.91\) cubic units

- Set the functions equal to find the limits. - Use the logarithmic antiderivative for \(1/x\). - For the volume, subtract the squares of the radii. - Rewrite \(1/x^2\) as a negative power before integrating.

1. Solving \(4-x=\frac{3}{x}\) gives \(x^2-4x+3=0\), so the curves intersect at \(x=1\) and \(x=3\). 2. On \([1,3]\), \(h(x)\ge k(x)\). Thus, \(A=\int_1^3\left(4-x-\frac{3}{x}\right)\,\mathrm{d}x\). 3. \(A=\left[4x-\frac{x^2}{2}-3\ln x\right]_1^3=4-3\ln3\approx0.704\) square units. 4. Using washers, \(V=\pi\int_1^3\left((4-x)^2-\left(\frac{3}{x}\right)^2\right)\,\mathrm{d}x\). 5. \(V=\pi\left[16x-4x^2+\frac{x^3}{3}+\frac{9}{x}\right]_1^3=\frac{8\pi}{3}\approx8.378\) cubic units.

a) \(4-3\ln3\) square units, or approximately \(0.704\) square units b) \(\frac{8\pi}{3}\) cubic units, or approximately \(8.378\) cubic units

- Find the intersection points first. - Identify the outer and inner radii. - Expand both squared binomials carefully. - Symmetry can shorten the calculation.

1. Solving \(x^2+1=9-x^2\) gives \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(g(x)\ge f(x)>0\), so washers have outer radius \(g(x)\) and inner radius \(f(x)\). 3. \(V=\pi\int_{-2}^{2}\left((9-x^2)^2-(x^2+1)^2\right)\,\mathrm{d}x\). 4. The integrand simplifies to \(80-20x^2\). 5. Thus, \(V=\pi\left[80x-\frac{20x^3}{3}\right]_{-2}^{2}=\frac{640\pi}{3}\approx670.21\) cubic units.

\(\frac{640\pi}{3}\) cubic units, or approximately \(670.21\) cubic units

- Set the function values equal and check solutions introduced by squaring. - The upper graph gives the outer radius on the enclosed interval. - Subtract the squares of the radii before integrating.

1. Set the functions equal: \(\sqrt{2x}=x\). Squaring and checking gives \(x=0\) and \(x=2\), so the intersection points are \((0,0)\) and \((2,2)\). 2. On \([0,2]\), \(f(x)\ge g(x)\). Using washers, \(V=\pi\int_0^2(2x-x^2)\,\mathrm{d}x\). 3. Therefore, \(V=\pi\left[x^2-\frac{x^3}{3}\right]_0^2=\frac{4\pi}{3}\) cubic units.

a) \((0,0)\) and \((2,2)\) b) \(\frac{4\pi}{3}\) cubic units

- Use the inner profile to find the cavity volume. - The delayed start of the inner profile creates a solid glass base. - Subtract the cavity volume from the full outer volume, then use mass equals density times volume.

1. The cavity volume is \(V_{\text{cavity}}=\pi\int_1^6(4x-4)\,\mathrm{d}x=50\pi\approx157.1\,\text{cm}^3=157.1\,\text{mL}\). 2. The full outer volume is \(V_{\text{outer}}=\pi\int_0^6(0.5x+3)^2\,\mathrm{d}x=126\pi\,\text{cm}^3\). 3. Therefore, the glass volume is \(76\pi\approx238.8\,\text{cm}^3\), and the mass is \(2.5(76\pi)=190\pi\approx596.9\,\text{g}\).

a) \(157.1\,\text{mL}\) b) \(76\pi\,\text{cm}^3\approx238.8\,\text{cm}^3\) c) \(596.9\,\text{g}\)

- Evaluate the radius function at both endpoints and at the midpoint. - Use the disc-method formula for the solid spool. - Convert the hole diameter to a radius, then subtract the cylindrical hole volume.

1. At both ends, \(f(0)=f(8)=\sqrt{12}\), so each end diameter is \(4\sqrt3\approx6.93\,\text{cm}\). At \(x=4\), the radius is \(2\), so the middle diameter is \(4\,\text{cm}\). 2. By the disc method, \(V_{\text{solid}}=\pi\int_0^8[0.5(x-4)^2+4]\,\mathrm{d}x=\frac{160\pi}{3}\approx167.55\,\text{cm}^3\). 3. The axial hole has radius \(1\,\text{cm}\) and length \(8\,\text{cm}\), so its volume is \(8\pi\,\text{cm}^3\). 4. Therefore, the wood volume is \(\frac{136\pi}{3}\approx142.42\,\text{cm}^3\).

a) End diameters: \(4\sqrt3\,\text{cm}\approx6.93\,\text{cm}\); middle diameter: \(4\,\text{cm}\) b) \(\frac{160\pi}{3}\,\text{cm}^3\approx167.55\,\text{cm}^3\) c) \(\frac{136\pi}{3}\,\text{cm}^3\approx142.42\,\text{cm}^3\)

- Use the intersections as the integration limits. - The quadratic is the outer radius on \([0,1]\). - Square both radius functions before subtracting.

1. The curves intersect at \(x=0\) and \(x=1\). 2. On \([0,1]\), \(x^2\ge x^4\). Thus, \(V=\pi\int_0^1(x^4-x^8)\,\mathrm{d}x\). 3. Therefore, \(V=\frac{4\pi}{45}\approx0.279\) cubic units.

\(\frac{4\pi}{45}\) cubic units, or approximately \(0.279\) cubic units

- Use the intersections as the integration limits. - The line is the outer radius on the enclosed interval. - Square both the coefficient and the variable in \(2x\).

1. The curves intersect at \(x=0\) and \(x=2\). 2. On \([0,2]\), \(2x\ge x^2\). Thus, \(V=\pi\int_0^2(4x^2-x^4)\,\mathrm{d}x\). 3. Therefore, \(V=\frac{64\pi}{15}\approx13.40\) cubic units.

\(\frac{64\pi}{15}\) cubic units, or approximately \(13.40\) cubic units

- Use the intersections as the integration limits. - The parabola gives the outer radius and the horizontal line gives the inner radius. - Expand the squared binomial, then use even symmetry.

1. The curves intersect at \(x=-2\) and \(x=2\). 2. On \([-2,2]\), \(f(x)\) is the outer radius and \(1\) is the inner radius. Thus, \(V=\pi\int_{-2}^{2}[(-x^2+5)^2-1]\,\mathrm{d}x\). 3. Expanding and using symmetry gives \(V=2\pi\int_0^2(x^4-10x^2+24)\,\mathrm{d}x=\frac{832\pi}{15}\approx174.25\) cubic units.

\(\frac{832\pi}{15}\) cubic units, or approximately \(174.25\) cubic units

- Use the inner profile for capacity and the outer profile for total volume. - The interval before the inner profile begins forms the solid base. - Subtract the cavity volume from the outer volume, then multiply by density.

1. The capacity is \(V_{\text{inside}}=\pi\int_1^4(2\sqrt{x-1})^2\,\mathrm{d}x=18\pi\approx56.55\,\text{cm}^3\). 2. The total outer volume is \(V_{\text{outer}}=\pi\int_0^4(0.2x+3)^2\,\mathrm{d}x=\frac{3484\pi}{75}\approx145.94\,\text{cm}^3\). 3. The glass volume is \(\frac{2134\pi}{75}\approx89.39\,\text{cm}^3\), so the mass is \(2.5\cdot\frac{2134\pi}{75}=\frac{1067\pi}{15}\approx223.47\,\text{g}\). 4. At the top, \(f(4)-g(4)=3.8-2\sqrt3\approx0.34\,\text{cm}\), the wall thickness at the rim.

a) \(18\pi\,\text{cm}^3\approx56.55\,\text{cm}^3\) b) \(\frac{2134\pi}{75}\,\text{cm}^3\approx89.39\,\text{cm}^3\) c) \(\frac{1067\pi}{15}\,\text{g}\approx223.47\,\text{g}\) d) It is the wall thickness at the top rim, approximately \(0.34\,\text{cm}\).

- Use horizontal distances from the y-axis as the washer radii. - The line is the outer boundary and the sideways parabola is the inner boundary. - Subtract the squared inner radius from the squared outer radius.

1. A horizontal slice has outer radius \(R(y)=5-y\) and inner radius \(r(y)=\frac{y^2}{4}\). 2. Therefore, \(V=\pi\int_0^2\left[(5-y)^2-\left(\frac{y^2}{4}\right)^2\right]\,\mathrm{d}y\). 3. Evaluating gives \(V=\frac{484\pi}{15}\) cubic units.

\(\frac{484\pi}{15}\) cubic units

- Measure both radii from the shifted axis, not from the x-axis. - The upper line is farther from the axis and gives the outer radius. - Subtract the squared distances before integrating.

1. The outer radius is \(R(x)=(x+1)-(-2)=x+3\), and the inner radius is \(r(x)=x-(-2)=x+2\). 2. Therefore, \(V=\pi\int_0^3[(x+3)^2-(x+2)^2]\,\mathrm{d}x=24\pi\) cubic units.

\(24\pi\) cubic units

- Translate the constant boundaries into constant radii. - Use the interval restriction to select the valid root.

1. The washers have outer radius \(5\) and inner radius \(k\). 2. \(V=\pi\int_0^4(25-k^2)\,\mathrm{d}x=4\pi(25-k^2)\). 3. Set \(4\pi(25-k^2)=64\pi\), so \(k^2=9\). 4. The restriction gives \(k=3\).

\(k=3\)

- Keep the unknown in the upper integration limit. - Simplify the difference of squared radii before integrating.

1. \(V=\pi\int_0^b[(2x)^2-x^2] \,\mathrm{d}x=\pi b^3\). 2. Set \(\pi b^3=21\pi\). 3. Thus \(b=\sqrt[3]{21}\).

\(b=\sqrt[3]{21}\)

- Compare the two pairs of distances to the axes. - A reflected placement can preserve both washer radii.

1. About \(y=0\), the washer area is \(\pi(5^2-2^2)=21\pi\). 2. About \(y=7\), the distances are \(5\) and \(2\) again, so the washer area is also \(21\pi\). 3. Both solids have the same length \(6\), so their volumes are equal.

The volumes are equal.

- Confirm which graph is farther from the axis on the whole interval. - Keep the difference of squared radii inside the numerical integral.

1. The upper curve gives the outer radius on \([0,1]\). 2. \(V=\pi\int_0^1\left[(2+e^{-x^2})^2-\left(1+\frac{x}{2}\right)^2\right] \,\mathrm{d}x\). 3. Numerical evaluation gives \(V\approx18.856\).

\(\approx18.856\) cubic units

- For an axis above the region, the lower boundary can be farther away. - Use symmetry only after forming the washer area correctly.

1. The outer radius is the distance from \(y=3\) to \(y=0\), so \(R=3\). 2. The inner radius is \(r(x)=3-x^2\). 3. \(V=\pi\int_{-1}^1[9-(3-x^2)^2] \,\mathrm{d}x=\frac{18\pi}{5}\).

\(\frac{18\pi}{5}\) cubic units

- Rewrite both boundaries as x-values in terms of \(y\). - Use the intersection heights as the integration limits. - Compare the two horizontal distances from the y-axis.

1. The curves meet at heights \(y=0\) and \(y=4\). Rewrite them as \(x=\sqrt{y}\) and \(x=\frac{y}{2}\). 2. A horizontal slice has outer radius \(\sqrt{y}\) and inner radius \(\frac{y}{2}\). 3. Therefore, \(V=\pi\int_0^4\left(y-\frac{y^2}{4}\right)\,\mathrm{d}y=\frac{8\pi}{3}\) cubic units.

\(\frac{8\pi}{3}\) cubic units

- The two radii share the same variable factor. - Use the stated range to select the physically valid parameter.

1. The outer radius is \(4x\), and the inner radius is \(kx\). 2. \(V=\pi\int_0^1(16-k^2)x^2\,\mathrm{d}x=\frac{\pi}{3}(16-k^2)\). 3. Set \(\frac{\pi}{3}(16-k^2)=4\pi\), giving \(k^2=4\). 4. The restriction gives \(k=2\).

\(k=2\)

- Write the area of a washer as the area of the outer circle minus the inner circle. - Compare that expression with the expansion of \((a-b)^2\). - Under what conditions are \(a^2-b^2\) and \((a-b)^2\) equal? - What does an inner radius of zero mean geometrically?

1. The actual washer area is \(A(x)=\pi(f(x)^2-g(x)^2)\). 2. The proposed expression is \(\pi(f(x)-g(x))^2=\pi(f(x)^2-2f(x)g(x)+g(x)^2)\). 3. Setting the expressions equal and simplifying gives \(2g(x)(f(x)-g(x))=0\). 4. Therefore, equality holds only when \(g(x)=0\), so the cross-section is a disc, or when \(f(x)=g(x)\), so the cross-sectional area is zero. 5. For a genuine washer with \(f(x)>g(x)>0\), the correct integrand is \(\pi(f(x)^2-g(x)^2)\). Squaring the radial thickness \(f(x)-g(x)\) ignores the washer's distance from the axis.

\(\pi(f^2-g^2)-\pi(f-g)^2=2\pi g(f-g)\). This difference is positive when \(f>g>0\), so \(\pi(f-g)^2\) cannot be used for a genuine washer. The correct volume uses \(\pi\int(f^2-g^2)\,\mathrm{d}x\).

- Read axial dimensions from x-values and radial dimensions from function values. - The delayed start of the interior profile creates the solid base. - Find material volume by subtracting the interior volume from the exterior volume.

1. The exterior extends from \(x=0\) to \(x=60\), so the tower's height is \(60\,\text{m}\). 2. The zeros are \(0\) and \(0.8\), so their difference is \(0.8\,\text{m}\). This is the axial thickness of the base. 3. At the top, \(f(60)-g(60)=\sqrt{270}-\sqrt{248.64}\approx0.66\,\text{m}\), the wall thickness there. 4. The gross volume is \(V_{\text{out}}=\pi\int_0^{60}4.5x\,\mathrm{d}x=8100\pi\approx25{,}446.90\,\text{m}^3\). 5. The interior volume is \(V_{\text{in}}=\pi\int_{0.8}^{60}4.2(x-0.8)\,\mathrm{d}x=7359.744\pi\,\text{m}^3\). 6. Therefore, the material volume is \(V_{\text{out}}-V_{\text{in}}=740.256\pi\approx2325.58\,\text{m}^3\).

a) \(60\,\text{m}\) b) \(0.8\,\text{m}\); the base thickness c) Approximately \(0.66\,\text{m}\); the wall thickness at the top d) \(8100\pi\,\text{m}^3\approx25{,}446.90\,\text{m}^3\) e) \(740.256\pi\,\text{m}^3\approx2325.58\,\text{m}^3\)

- Use horizontal slices because the axis of rotation is vertical. - For an axis to the right, the leftmost curve gives the outer radius. - Write each radius as a positive horizontal distance from \(x=6\).

1. The axis is to the right of both curves. The left boundary is farther from the axis, so \(R(y)=6-y^2\). 2. The inner radius is \(r(y)=6-(2y+1)=5-2y\). 3. Therefore, \(V=\pi\int_0^1[(6-y^2)^2-(5-2y)^2]\,\mathrm{d}y=\frac{238\pi}{15}\) cubic units.

\(\frac{238\pi}{15}\) cubic units

- Use the figure to check whether each vertical slice stays on one side of the axis. - A slice that crosses the axis produces a disc with no hole. - Split the integral where the lower boundary crosses the x-axis.

1. On \([0,1]\), the vertical slice crosses the x-axis, so rotating it produces a disc of radius \(2\), not a washer with an inner radius. 2. On \([1,2]\), the slice remains above the x-axis and produces a washer with outer radius \(2\) and inner radius \(x-1\). 3. Thus, \(V=\pi\int_0^1 4\,\mathrm{d}x+\pi\int_1^2[4-(x-1)^2]\,\mathrm{d}x=\frac{23\pi}{3}\). 4. The proposed value is \(V_s=\frac{22\pi}{3}\), so it underestimates the volume by \(\frac{\pi}{3}\).

The proposal is incorrect. The correct volume is \(\frac{23\pi}{3}\) cubic units, and \(V_s\) is too small by \(\frac{\pi}{3}\) cubic units.

- Check whether the vertical axis lies inside each horizontal slice. - The cross-sectional type changes when the left boundary reaches \(x=1\). - Measure every radius as a horizontal distance from the dashed axis.

1. For \(0\le y\le1\), the horizontal slice crosses \(x=1\), producing a disc of radius \(3-y\). 2. For \(1\le y\le2\), both boundaries lie to the right of the axis, producing a washer with outer radius \(3-y\) and inner radius \(y-1\). 3. Thus, \(V=\pi\int_0^1(3-y)^2\,\mathrm{d}y+\pi\int_1^2[(3-y)^2-(y-1)^2]\,\mathrm{d}y\). 4. The two contributions are \(\frac{19\pi}{3}\) and \(2\pi\), so \(V=\frac{25\pi}{3}\) cubic units.

\(\frac{25\pi}{3}\) cubic units