# Surface Area of a Solid of Revolution

#### hyperbola

Given $$\displaystyle y = x^3$$

$$\displaystyle SA = 2\pi\int y\cdot ds$$

$$\displaystyle ds = \sqrt{1+y'^2}=\sqrt{1+(3x^2)^2}=\sqrt{1+9x^4}$$

The solid of revolution is rotated about the x-axis. Limits of integration are taken to be a = 0 and b = 1.

$$\displaystyle SA = 2\pi{\int^1_0} x^3 \sqrt{1+9x^4}\cdot dx$$

Let $$\displaystyle u = 1+9x^4$$, $$\displaystyle du = 36x^3\cdot dx$$

New limits of integration:

$$\displaystyle b = \sqrt{1+9(1)^4} = \sqrt{10}$$
$$\displaystyle a = \sqrt{1+9(0)^4} = 1$$

$$\displaystyle SA = \frac{1}{18}\pi\int^{\sqrt{10}}_1 u\cdot du$$

$$\displaystyle SA = \frac{1}{18}\pi \left[\frac{u^2}{2}\right]^{\sqrt{10}}_1$$

Substitute for u... Is this correct? Thank you in advance.

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Looks like textbook example to me

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#### v8archie

Math Team
I don't see any problems with that. (Apart from the missing $\mathrm dx$ differentials in the working for $\mathrm ds$).

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#### Math Message Board tutor

$$\displaystyle SA = \frac{1}{18}\pi \left[\frac{u^2}{2}\right]^{\sqrt{10}}_1$$

Substitute for u... Is this correct? Thank you in advance.
Before substituting into u, you should simplify it first:

$$\displaystyle SA \ = \ \frac{\pi}{36} (u^2)\bigg|^{\sqrt{10}}_1$$

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#### v8archie

Math Team
"Could" rather than "should" in my opinion.