Solutions to Problem Set 12

Math 211-03

10-2-2017

[Volumes of revolution]

1. Let R be the region in the first quadrant cut off by the line $3 x + 4 y = 12$ . Find the volume of the solid generated when R is revolved about the x-axis.

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution1.eps}}$$

Use circular slices. The radius of a typical slice is $r = y = \dfrac{12 - 3 x}{4}$ . The thickness of a typical slice is $dx$ . The volume is

$$\int_0^4 \pi \left(\dfrac{12 - 3 x}{4}\right)^2\,dx = \dfrac{\pi}{16} \int_0^4 (144 - 72 x + 9 x^2)\,dx = \dfrac{\pi}{16} \left[144 x - 36 x^2 + 3 x^3\right]_0^4 = 12 \pi = 37.69911 \ldots.\quad\halmos$$


2. Let R be the area bounded by $y = \sin x$ and the x-axis, from $x = 0$ to $x = \pi$ . Find the volume of the solid generated when R is revolved about the x-axis.

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution2.eps}}$$

Use circular slices. The radius of a typical slice is $r = y = \sin x$ . The thickness of a typical slice is $dx$ . The volume is

$$\int_0^\pi \pi (\sin x)^2\,dx = \pi \int_0^\pi \dfrac{1}{2} (1 - \cos 2 x)\,dx = \dfrac{\pi}{2} \left[x - \dfrac{1}{2} \sin 2 x\right]_0^\pi = \dfrac{\pi^2}{2} = 4.93480 \ldots.\quad\halmos$$


3. Let R be the area in the first quadrant bounded above by $y = 1$ and below by $y
   = x^2$ . Find the volume of the solid generated when R is revolved about the y-axis.

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution3.eps}}$$

Use circular slices. The radius of a typical slice is $r = x = \sqrt{y}$ . The thickness of a typical slice is $dy$ . The volume is

$$\int_0^1 \pi (\sqrt{y})^2\,dy = \pi \int_0^1 y\,dy = \pi \left[\dfrac{1}{2} y^2\right]_0^1 = \dfrac{\pi}{2} = 1.57079 \ldots.\quad\halmos$$


4. Let R be the region between $y = x^2 + 1$ and the x-axis, from $x = 0$ to $x
   = 1$ . Find the volume generated by revolving R about the x-axis.

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution4.eps}}$$

Use circular slices. The radius of a typical slice is $r = y = x^2 + 1$ . The thickness of a typical slice is $dx$ . The volume is

$$\int_0^1 \pi (x^2 + 1)^2\,dx = \pi \int_0^1 \pi (x^4 + 2 x^2 + 1)\,dx = \pi \left[\dfrac{1}{5} x^5 + \dfrac{2}{3} x^3 + x\right]_0^1 = \dfrac{28 \pi}{15} = 5.86430 \ldots.\quad\halmos$$


5. Let R be the region between $y = x$ and $y = 4 x$ , from $x = 0$ to $x = 1$ . Find the volume of the solid generated when R is revolved about the x-axis.

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution7.eps}}$$

Use washers. The inner radius of a typical washer is $r_{\rm in} = x$ . The outer radius of a typical washer is $r_{\rm out} = 4 x$ . The thickness of a typical washer is $dx$ . The volume is

$$\int_0^1 \pi \left((4 x)^2 - x^2\right)\,dx = \pi \int_0^1 15 x^2\,dx = \pi \left[5 x^3\right]_0^1 = 5 \pi = 15.70796 \ldots.\quad\halmos$$


6. Let R be the region in the first quadrant cut off by the line $3 x + 4 y = 12$ . Find the volume of the solid generated when R is revolved about the line $y = 5$ .

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution5.eps}}$$

Use washers. The inner radius of a typical washer is $r_{\rm in} = 5 - y = 5 -
   \dfrac{1}{4} (12 - 3 x) = 2 + \dfrac{3}{4} x$ . The outer radius of a typical washer is $r_{\rm out} = 5$ . The thickness of a typical washer is $dx$ . The volume is

$$\int_0^4 \pi \left(5^2 - \left(2 + \dfrac{3}{4} x\right)^2\right)\,dx = \pi \int_0^4 \left(21 - 3 x - \dfrac{9}{16} x^2\right)\,dx =$$

$$\pi \left[21 x - \dfrac{3}{2} x^2 - \dfrac{3}{16} x^3\right]_0^4 = 48 \pi = 150.79644 \ldots.\quad\halmos$$


He who walks in another's tracks leaves no footprints. - Joan Brannon


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