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anonymous
 one year ago
Find the exact area of the surface obtained by rotating the curve about the xaxis.
y =( (x^3)/3)+(1/4x)
1/2 <= x <= 1
Need Help!
anonymous
 one year ago
Find the exact area of the surface obtained by rotating the curve about the xaxis. y =( (x^3)/3)+(1/4x) 1/2 <= x <= 1 Need Help!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A solution using the Mathematica computer program is attached.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@robtobey: You probably misread the question. Not the volume of the body is needed, but its surface area. The formula for the surface area is: \(\int_{0.5}^1 2\pi f(x)\sqrt{1+(f'(x))^2}dx \) This would be: \( \int_{0.5}^1 2\pi(\frac{x^3}{3}+\frac{x}{4})\sqrt{1+(x^2+\frac{1}{4})^2}dx\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not confident if I can do this... Looking it up in WolframAlpha confirms my fears:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Refer to the attachment from the Mathematica v9 program.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ZeHanz the integral isn't too bad, as it turns out: \[2\pi\int_{1/2}^1 \left(\frac{x^3}{3}+\frac{x}{4}\right)\sqrt{1+\left(x^2+\frac{1}{4}\right)^2}\,dx\] Pull out a factor of \(\frac{x}{3}\), rewrite, then substitute \(t=x^2+\dfrac{1}{4}\) so that \(\dfrac{1}{2}\,dt=x\,dx\), giving \[\frac{2\pi}{3}\int_{1/2}^1 x\left(x^2+\frac{3}{4}\right)\sqrt{1+\left(x^2+\frac{1}{4}\right)^2}\,dx=\frac{\pi}{3}\int_{1/2}^{5/4}\left(t+\frac{1}{2}\right)\sqrt{1+t^2}\,dt\] Next, set \(t=\tan v\), so \(dt=\sec^2v\,dv\), then the integral becomes \[\frac{\pi}{3}\int_{\arctan(1/2)}^{\arctan(5/4)}\left(\tan v+\frac{1}{2}\right)\sqrt{1+\tan^2v}\,\sec^2v\,dv\] Simplifying, \[\frac{\pi}{3}\int_{\arctan(1/2)}^{\arctan(5/4)}\left(\tan v+\frac{1}{2}\right)\sec^3v\,dv\] Upon expanding, we can easily deal with the remaining integration. \[\begin{align*}\int \tan v\sec^3v\,dv&=\int \tan v\sec v\sec^2v\,dv\\[1ex]&=\int \sec^2v\,d(\sec v)\\[1ex]&=\frac{1}{3}\sec v+C\\[2ex]\hline \int\sec^3v\,dv&=\sec v\tan v\int \sec v\tan^2v\,dv\\[1ex] &=\sec v\tan v\int \frac{\sin^2v}{\cos^3v}\,dv\\[1ex] &=\sec v\tan v\int \frac{1\cos^2v}{\cos^3v}\,dv\\[1ex] &=\sec v\tan v\int (\sec^3v\sec v)\,dv\\[1ex] 2\int\sec^3v\,dv&=\sec v\tan v+\int \sec v\,dv\\[1ex] \int\sec^3v\,dv&=\frac{\sec v\tan v+\ln\sec v+\tan v}{2}+C\end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Typo, that first antiderivative should be \(\dfrac{1}{3}\sec^{\color{red}3}v+C\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, that's impressive stuff, imo!
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