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

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1. anonymous

A solution using the Mathematica computer program is attached.

2. anonymous

@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: $$\int_{0.5}^1 2\pi(\frac{x^3}{3}+\frac{x}{4})\sqrt{1+(x^2+\frac{1}{4})^2}dx$$

3. anonymous

I'm not confident if I can do this... Looking it up in WolframAlpha confirms my fears:

4. anonymous

Refer to the attachment from the Mathematica v9 program.

5. anonymous

@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*}

6. anonymous

Typo, that first antiderivative should be $$\dfrac{1}{3}\sec^{\color{red}3}v+C$$

7. anonymous

Well, that's impressive stuff, imo!