Surface Area of a Solid of Revolution

Dec 2014
142
6
The Asymptote
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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v8archie

Math Team
Dec 2013
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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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Jun 2014
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\(\displaystyle SA = \frac{1}{18}\pi \left[\frac{u^2}{2}\right]^{\sqrt{10}}_1\)

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
Dec 2013
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2,682
Colombia
"Could" rather than "should" in my opinion.