Solutions to Problem Set 13

Math 211-03

10-3-2017

[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 line $y = -3$ .

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution6.eps}}$$

Use washers. The inner radius of a typical washer is $r_{\rm in} = 3$ .

The outer radius of a typical washer is

$$r_{\rm out} = 3 + \dfrac{1}{4} (12 - 3 x) = 6 - \dfrac{3}{4} x.$$

The thickness of a typical washer is $dx$ . The volume is

$$\int_0^4 \pi \left(\left(6 - \dfrac{3}{4} x\right)^2 - 3^2\right)\,dx = \pi \int_0^4 \left(\dfrac{9}{16} x^2 - 9 x + 27\right)\,dx =$$

$$\pi \left[\dfrac{3}{16} x^3 - \dfrac{9}{2} x^2 + 27 x\right]_0^4 = 48 \pi = 150.79644 \ldots.\quad\halmos$$


2. Let R be the region inside the square defined by the inequalities $1 \le x \le 2$ and $0
   \le y \le 1$ .

$$\hbox{\epsfysize=1.5in \epsffile{volumes-of-revolution8a.eps}}$$

Find the volume generated when R is revolved about the y-axis.

$$\hbox{\epsfysize=1.5in \epsffile{volumes-of-revolution8b.eps}}$$

The inner radius is $r_{in} =
   1$ and the outer radius is $r_{out} = 2$ . The thickness of a typical washer is $dy$ . The volume is

$$\int_0^1 \pi (2^2 - 1^2)\,dy = \pi \int_0^1 3\,y = \pi \left[3 y\right]_0^1 = 3 \pi = 9.42477 \ldots.\quad\halmos$$


3. 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 $x = 5$ .

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution9.eps}}$$

Use cylindrical shells. The radius of a typical shell is $r = 5 - x$ , the height of a typical shell is $h = \dfrac{12 - 3 x}{4}$ , and the thickness is $dx$ . The volume is

$$\int_0^4 2 \pi (5 - x)\left(\dfrac{12 - 3 x}{4}\right)\,dx = \dfrac{\pi}{2} \int_0^4 (5 - x)(12 - 3 x)\,dx = \dfrac{\pi}{2} \int_0^4 (3 x^2 - 27 x + 60)\,dx =$$

$$\dfrac{\pi}{2} \left[x^3 - \dfrac{27}{2}x^2 + 60 x\right]_0^4 = 44 \pi \approx 138.23008.\quad\halmos$$

Note: It's possible to do this problem using washers.


4. 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 $x =
   -2$ .

$$\hbox{\epsfysize=1.75in \epsffile{volumes-of-revolution10.eps}}$$

Use cylindrical shells. The radius of a typical shell is $r = x + 2$ , the height of a typical shell is $h = \dfrac{12 - 3 x}{4}$ , and the thickness is $dx$ . The volume is

$$\int_0^4 2 \pi (x + 2)\left(\dfrac{12 - 3 x}{4}\right)\,dx = \dfrac{\pi}{2} \int_0^4 (x + 2)(12 - 3 x)\,dx = \dfrac{\pi}{2} \int_0^4 (24 + 6 x - 3 x^2)\,dx =$$

$$\dfrac{\pi}{2} \left[24 x + 3 x^2 - x^3\right]_0^4 = 40 \pi \approx 125.66371.\quad\halmos$$

Note: It's possible to do this problem using washers.


5. Let R be the region bounded by the graph of $y = \sin x$ and the x-axis, from $x = 0$ to $x = \pi$ . Find the volume generated by revolving R about the y-axis.

$$\hbox{\epsfysize=1.5in \epsffile{volumes-of-revolution11.eps}}$$

It's possible to do this problem using washers, but it would be messier. Instead, use cylindrical shells. The radius of a typical shell is $r = x$ , the height of a typical shell is $h = \sin x$ , and the thickness is $dx$ . The volume is

$$\int_0^\pi 2 \pi x \sin x\,dx = 2 \pi \left[-x \cos x + \sin x\right]_0^\pi = 2 \pi^2 = 19.73920 \ldots.$$

Here's the work for the integral, which is done by parts:

$$\matrix{ & \displaystyle \der {} x & \displaystyle \int\,dx \cr \noalign{\vskip2pt} + & x & \sin x \cr \noalign{\vskip2pt} - & 1 & -\cos x \cr \noalign{\vskip2pt} + & 0 & -\sin x \cr}$$

$$\int x \sin x\,dx = -x \cos x + \sin x + c.\quad\halmos$$


6. Let R be the region bounded by the curves $y = x^2$ and $y = 12 - 2
   x^2$ . Find the volume generated by revolving R about the line $x
   = 3$ .

$$\hbox{\epsfysize=1.5in \epsffile{volumes-of-revolution12.eps}}$$

Find the intersection points of the curves:

$$\eqalign{ x^2 & = 12 - 2 x^2 \cr 3 x^2 & = 12 \cr x^2 & = 4 \cr x & = \pm 2 \cr}$$

Use cylindrical shells.The radius of a typical shell is $r = 3 - x$ , the height of a typical shell is

$$h = 12 - 2 x^2 - x^2 = 12 - 3 x^2.$$

The thickness is $dx$ . The volume is

$$\int_{-2}^2 2 \pi (3 - x)(12 - 3 x^2)\,dx = 2 \pi \int_{-2}^2 \left(3 x^3 - 9 x^2 - 12 x + 36\right)\,dx =$$

$$2 \pi \left[\dfrac{3}{4} x^4 - 3 x^3 - 6 x^2 + 36 x\right]_{-2}^2 = 192 \pi = 603.18578 \ldots.\quad\halmos$$

Note: You could do this problem using washers, but it would require 2 integrals. Do you see why?


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