Question

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!

Answer 1

A solution using the Mathematica computer program is attached.

Answer 2

@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:

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

Answer 3

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

Answer 4

Refer to the attachment from the Mathematica v9 program.

Answer 5

@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.
\[\begin{align*}\int \tan v\sec^3v\,dv&=\int \tan v\sec v\sec^2v\,dv\\[1ex]&=\int \sec^2v\,d(\sec v)\\[1ex]&=\frac{1}{3}\sec v+C\\[2ex]\hline
\int\sec^3v\,dv&=\sec v\tan v-\int \sec v\tan^2v\,dv\\[1ex]
&=\sec v\tan v-\int \frac{\sin^2v}{\cos^3v}\,dv\\[1ex]
&=\sec v\tan v-\int \frac{1-\cos^2v}{\cos^3v}\,dv\\[1ex]
&=\sec v\tan v-\int (\sec^3v-\sec v)\,dv\\[1ex]
2\int\sec^3v\,dv&=\sec v\tan v+\int \sec v\,dv\\[1ex]
\int\sec^3v\,dv&=\frac{\sec v\tan v+\ln|\sec v+\tan v|}{2}+C\end{align*}\]

Answer 6

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

Answer 7

Well, that's impressive stuff, imo!