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Section 7.2 #24: Find the volume of the solid generated by revolving the region bounded by $y=x\sqrt{4-x^2}$ and $y=0$ about the $x$-axis.
Solution: Sketch the region and washers:
Note that we will use two integrals to compute the area of the solid of revolution. This is for the same reason we would have to use two integrals to compute the area of the region: the "top" function and the "bottom" function change after the intersection point. Therefore compute $$\begin{array}{ll} \mathrm{Volume} &= \pi\displaystyle\int_{-2}^0 \left[x \sqrt{4-x^2} \right]^2 \mathrm{d}x + \pi\displaystyle\int_0^2 \left[x \sqrt{4-x^2} \right]^2 \mathrm{d}x \\ &= \pi \displaystyle\int_{-2}^0 4x^2-x^4 \mathrm{d}x + \pi \displaystyle\int_0^2 4x^2-x^4 \mathrm{d}x \\ &=\pi \left[ \dfrac{4}{3} x^3 - \dfrac{1}{5}x^5 \right|_{-2}^0 + \pi \left[ \dfrac{4}{3} x^3 - \dfrac{1}{5} x^5 \right|_0^2 \\ &=\pi \left\{ 0 - \left[ -\dfrac{32}{3}+ \dfrac{32}{5} \right] \right\} + \pi \left\{ \left[ \dfrac{32}{3} - \dfrac{32}{5} \right] - 0 \right\} \\ &=\pi \left[ \dfrac{64}{3} - \dfrac{64}{5} \right] \\ &= \dfrac{128\pi}{15}. \end{array}$$

Section 7.2 #28: Find the volume of the solid generated by revolving the region bounded by $y=e^{\frac{x}{4}}, y=0, x=0$, and $x=6$ about the $x$-axis.
Solution: Sketch the region:
Now sketch the revolved solid and a washer:
The radius of the washer is $R(x)=e^{\frac{x}{4}}$, so we can now compute the volume. Compute $$\begin{array}{ll} \mathrm{Volume} &= \pi \displaystyle\int_0^6 \left( e^{\frac{x}{4}} \right)^2 \mathrm{d}x \\ &= \pi \displaystyle\int_0^6 e^{\frac{x}{2}} \mathrm{d}x \\ &\stackrel{u=\frac{x}{2}}{=} 2\pi \displaystyle\int_0^3 e^u \mathrm{d}u \\ &= 2\pi \Bigg[ e^u \Bigg|_0^3 \\ &= 2\pi(e^3-1). \end{array}$$

Section 7.2 #33: Find the volume of the solid generated by revolving the region bounded by the graphs of $y=\sin(x), y=0, x=0,$ and $x=\pi$ about the $x$-axis.
Solution: First sketch the region:
Now sketched the revolved region and a washer:
Recall the trigonometric identity $$\sin^2(x) = \dfrac{1-\cos(2x)}{2}.$$ Now find the volume: compute $$\begin{array}{ll} \mathrm{Volume} &= \pi \displaystyle\int_0^{\pi} \sin^2(x) \mathrm{d}x \\ &=\pi \displaystyle\int_0^{\pi} \dfrac{1-\cos(x)}{2} \mathrm{d}x \\ &=\dfrac{\pi}{2} \displaystyle\int_0^{\pi} 1 \mathrm{d}x - \dfrac{\pi}{2} \displaystyle\int_0^{\pi} \cos(x) \mathrm{d}x \\ &= \dfrac{\pi}{2} \Bigg[ x \Bigg|_0^{\pi} - \dfrac{\pi}{2} \Bigg[ \sin(x) \Bigg|_0^{\pi} \\ &=\dfrac{\pi}{2}(\pi-0) - \dfrac{\pi}{2}(0-0) \\ &= \dfrac{\pi^2}{2}. \end{array}$$

Section 7.3 #8: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by $y=9-x^2$ and $y=0$ about the $y$-axis.
Solution: Draw the region:
and draw the shell:
Note that when this region rotates around the $x$-axis, the "right half" will intersect the path of the "left half". To avoid overcounting the volume, we will simply rotate the right half of the region instead. Now compute $$\begin{array}{ll} \mathrm{Volume} &= 2\pi \displaystyle\int_0^3 x (9-x^2) \mathrm{d}x \\ &= 2\pi \Bigg[ \dfrac{9}{2}x^2 - \dfrac{1}{4} x^4 \Bigg|_0^3 \\ &= 2\pi \left( \dfrac{81}{2} - \dfrac{81}{4} \right)\\ &=\dfrac{162\pi}{4} \\ &=\dfrac{81\pi}{2}. \end{array}$$

Section 7.3 #14: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by $y = \left\{ \begin{array}{ll} \dfrac{\sin(x)}{x},& \quad x>0 \\ 1,& \quad x=0 \end{array} \right\}, y=0, x=0,$ and $x=\pi$ about the $y$-axis.
Solution: First draw the region
and draw a shell
Therefore compute $$\begin{array}{ll} \mathrm{Volume} &= 2\pi \displaystyle\int_0^{\pi} x \left( \dfrac{\sin(x)}{x} \right) \mathrm{d}x \\ &= 2 \pi \displaystyle\int_0^{\pi} \sin(x) \mathrm{d}x \\ &= 2 \pi \Bigg[ -\cos(x) \Bigg|_0^{\pi} \\ &= 2 \pi [-(-1)-(-1)] \\ &= 4\pi. \end{array}$$

Section 7.3 #20: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by $y=4x^2$, $x=0$, and $y=4$ about the $x$-axis.
Solution: First note that the intersection point of $y=4$ and $y=4x^2$ is at $x=1$ with height $y=4$. Now draw the region
and draw a shell
Therefore calculate $$\begin{array}{ll} \mathrm{Volume} &= 2\pi \displaystyle\int_0^4 \dfrac{1}{2} y^{\frac{3}{2}} \mathrm{d}y \\ &=2\pi \left[ \dfrac{1}{5} y^{\frac{5}{2}} \right|_0^4 \\ &=\dfrac{64\pi}{5}. \end{array}$$ Section 7.3 #24: Use the shell method to find the volume of the solid generated by revolving the region bounded by $y=\sqrt{x}, y=0$, and $y=4$ about the line $x=6$.
Solution: First draw the region
and draw the shell
Now calculate $$\begin{array}{ll} \mathrm{Volume} &= 2\pi \displaystyle\int_0^4 \sqrt{x}(6-x) \mathrm{d}x \\ &= 12\pi\displaystyle\int_0^4 x^{\frac{1}{2}} \mathrm{d}x - 2\pi\displaystyle\int_0^4 x^{\frac{3}{2}} \mathrm{d}x \\ &=12\pi \Bigg[ \dfrac{2}{3} x^{\frac{3}{2}} \Bigg|_0^4 - 2\pi \Bigg[ \dfrac{2}{5} x^{\frac{5}{2}} \Bigg|_0^4 \\ &= \dfrac{24\pi}{3} [4^{\frac{3}{2}}-0] - \dfrac{4\pi}{5} [ 4^{\frac{5}{2}}-0] \\ &=64\pi - \dfrac{128\pi}{5} \\ &=\dfrac{320\pi}{5} - \dfrac{128\pi}{5} \\ &= \dfrac{192\pi}{5}. \end{array}$$

Section 7.3 #25: Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region bounded by $y=x^2$ and $y=4x-x^2$ about the line $x=4$.
Solution: First draw the region
and the shell
Now calculate $$\begin{array}{ll} \mathrm{Volume} &= 2\pi \displaystyle\int_0^2 (4-x)((4x-x^2)-x^2) \mathrm{d}x \\ &= 2\pi \displaystyle\int_0^2 (4-x)(4x-2x^2) \mathrm{d}x \\ &= 2\pi \displaystyle\int_0^2 2x^3-12x^2+16x \mathrm{d}x \\ &= 2\pi \Bigg[ \dfrac{2}{4}x^4 - \dfrac{12}{3} x^3 + \dfrac{16}{2} x^2 \Bigg|_0^2 \\ &= 2 \pi \Bigg[\dfrac{x^4}{2} - 4x^3 + 8x^2 \Bigg|_0^2 \\ &= 2\pi [(8 - 32 + 32)-0] \\ &= 16\pi. \end{array}$$