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November 12th, 2016, 06:15 PM   #1
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Math Focus: Certainty
Surface Area of a Solid of Revolution

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.

Last edited by skipjack; November 13th, 2016 at 01:54 AM.
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November 12th, 2016, 06:25 PM   #2
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Looks like textbook example to me
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November 12th, 2016, 06:26 PM   #3
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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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Last edited by greg1313; November 12th, 2016 at 06:50 PM.
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November 12th, 2016, 10:59 PM   #4
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Quote:
Originally Posted by hyperbola View Post

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

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

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

Last edited by skipjack; November 13th, 2016 at 01:50 AM.
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November 13th, 2016, 04:32 AM   #5
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"Could" rather than "should" in my opinion.
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