Solutions to Problem Set 33

Math 211-03

11-28-2017

[Surfaces of revolution]

1. Find the area of the surface generated by revolving the curve $y = \dfrac{1}{2} x^2 + 1$ for $0
   \le x \le 1$ about the y-axis.

$$\der y x = x, \quad 1 + \left(\der y x\right)^2 = 1 + x^2.$$

Since the curve is revolved about the y-axis, the radius is $R = x$ . The area is

$$\int_0^1 2 \pi x \sqrt{1 + x^2}\,dx = 2 \pi \left[\dfrac{1}{3} (1 + x^2)^{3/2}\right]_0^1 = \dfrac{2}{3} \pi (2^{3/2} - 1) = 3.82944 \ldots .\quad\halmos$$


2. Find the area of the surface generated by revolving $y = x^3$ from $x = 0$ to $x = 2$ about the x-axis.

$$\eqalign{ \der y x & = 3 x^2 \cr \left(\der y x\right)^2 & = 9 x^4 \cr \sqrt{1 + \left(\der y x\right)^2} & = \sqrt{1 + 9 x^4} \cr}$$

$R = y = x^3$ , so the area is

$$S = \int_0^2 2 \pi x^3 \sqrt{1 + 9 x^4}\,dx = 2 \pi \int_1^{145} x^3 \sqrt{u} \cdot \dfrac{du}{36 x^3} = \dfrac{\pi}{18} \int_1^{145} \sqrt{u}\,du =$$

$$\left[u = 1 + 9 x^4, \quad du = 36 x^3\,dx, \quad dx = \dfrac{du}{36 x^3}; \quad x = 0, \quad u = 1; \quad x = 2, \quad u = 145\right]$$

$$\dfrac{\pi}{18} \left[\dfrac{2}{3} u^{3/2}\right]_1^{145} = \dfrac{\pi}{27} (145^{3/2} - 1) = 203.04360 \ldots.\quad\halmos$$


3. Find the area of the surface generated by revolving $y = \dfrac{2}{3} x^{3/2} -
   \dfrac{1}{2} x^{1/2}$ , $1 \le x \le 4$ , about the x-axis.

$$\hbox{\epsfysize=1.5in \epsffile{surfaces-of-revolution1.eps}}$$

$$\eqalign{ \der y x & = x^{1/2} - \dfrac{1}{4 x^{1/2}} \cr \left(\der y x\right)^2 & = x - \dfrac{1}{2} + \dfrac{1}{16x} \cr 1 + \left(\der y x\right)^2 & = x + \dfrac{1}{2} + \dfrac{1}{16x} \cr}$$

Notice that this is just $\left(\der y x\right)^2$ with the sign of the middle term changed. But $\left(\der y x\right)^2$ was $x^{1/2} - \dfrac{1}{4 x^{1/2}}$ squared, so $1 +
   \left(\der y x\right)^2$ must be $x^{1/2} + \dfrac{1}{4
   x^{1/2}}$ squared:

$$1 + \left(\der y x\right)^2 = \left(x^{1/2} + \dfrac{1}{4 x^{1/2}}\right)^2.$$

Thus,

$$\sqrt{1 + \left(\der y x\right)^2} = x^{1/2} + \dfrac{1}{4 x^{1/2}}.$$

Now $R = y = \dfrac{2}{3}
   x^{3/2} - \dfrac{1}{2} x^{1/2}$ , so the area is

$$S = \int_1^4 2 \pi \left(\dfrac{2}{3} x^{3/2} - \dfrac{1}{2} x^{1/2}\right) \left(x^{1/2} + \dfrac{1}{4 x^{1/2}}\right)\,dx = 2 \pi \int_1^4 \left(\dfrac{2}{3} x^2 - \dfrac{1}{3} x -\dfrac{1}{8}\right)\,dx = 2 \pi \left[\dfrac{2 x^3}{9} - \dfrac{x^2}{6} - \dfrac{x}{8}\right]_1^4 =$$

$$\dfrac{89 \pi}{4} = 69.90043 \ldots.\quad\halmos$$


4. Find the area of the surface generated by revolving the following curve about the y-axis:

$$x = e^t + e^{-t}, \quad y = 2 t, \quad 0 \le t \le 1.$$

$$\der x t = e^t - e^{-t}, \quad \der y t = 2,$$

$$\left(\der x t\right)^2 = e^{2 t} - 2 + e^{-2 t}, \quad \left(\der y t\right)^2 = 4,$$

$$\left(\der x t\right)^2 + \left(\der y t\right)^2 = e^{2 t} + 2 + e^{-2 t} = (e^t + e^{-t})^2,$$

$$\sqrt{\left(\der x t\right)^2 + \left(\der y t\right)^2} = e^t + e^{-t}.$$

Since the curve is revolved about the y-axis, $R = x = e^t + e^{-t}$ . So the area is

$$\int_0^1 2 \pi (e^t + e^{-t})(e^t + e^{-t})\,dt = 2 \pi \int_0^1 (e^{2 t} + 2 + e^{-2 t})\,dt = 2 \pi \left[\dfrac{1}{2} e^{2 t} + 2 t - \dfrac{1}{2} e^{-2 t}\right]_0^1 =$$

$$\pi(e^2 - e^{-2} + 4) = 35.35460 \ldots .\quad\halmos$$


Strength doesn't come from physical capacity. It comes from indomitable will. - Mahatma Gandhi


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