Volume of solid bound by Trig function

[kneaiak]

"Use the disk method to find the volume of the solid generated when the region bound by y=3sinx and y=0, for 0<=x<=pi, is revolved about the x-axis. (Recall that (sin x) squared = 1/2 (1-cos2x)).)"

This is what I have so far:

int from 0 to pi of (pi)3sinx dx

(pi) int from 0 to pi of (3sinx)squared dx

(pi) int from 0 to pi of 9(sinx)squared dx

(pi) int from 0 to pi of 9(1/2)(1-cos2x) dx

Here's where I'm pretty much stuck. The below is what I finally ended up with but it doesn't make sense. Any help would be greatly appreciated.

(9pi)/2 x - (1/2)sin 2x evaluated from 0 to pi

 

[MarkFL]

The volume of an arbitrary disk is:

\(\displaystyle dV=\pi r^2\,dx\)

\(\displaystyle r=y=3\sin(x)\) and so we have:

\(\displaystyle dV=9\pi \sin^2(x)\,dx\)

Summing up the disks, we find:

\(\displaystyle \displaystyle V=9\pi\int_0^{\pi}\sin^2(x)\,dx=\frac{9\pi}{2}\int_0^{\pi} 1-\cos(2x)\,dx=\frac{9\pi}{2}\left[x-\frac{1}{2}\sin(2x) \right]_0^{\pi}\)

\(\displaystyle \displaystyle V=\frac{9\pi}{2}\left(\left(\pi-\frac{1}{2}\sin(2\pi) \right)-\left(0-\frac{1}{2}\sin(2\cdot0) \right) \right)=?\)

 

[kneaiak]

(18(pi)squared + 9(pi))/4

?

Is there a way I can enter problems the same way you're able to, with the integral symbol, etc?

 

[MarkFL]

No, we have:

\(\displaystyle \displaystyle V=\frac{9\pi}{2}\left(\left(\pi-\frac{1}{2}\sin(2\pi) \right)-\left(0-\frac{1}{2}\sin(2\cdot0) \right) \right)\)

Recall that \(\displaystyle \sin(0)=\sin(2\pi)=0\) and thus:

\(\displaystyle \displaystyle V=\frac{9\pi}{2}\left(\left(\pi-\frac{1}{2}\cdot0 \right)-\left(0-\frac{1}{2}\cdot0 \right) \right)\)

\(\displaystyle \displaystyle V=\frac{9\pi}{2}(\pi)=\frac{9\pi^2}{2}\)

To render your expressions as I have done requires the use of \(\displaystyle \LaTeX\). Basically you wrap valid LaTeX code in tex tags.

Do an online search of "using latex" and you will find many online tutorials on how to use it.

edit: another way to see how LaTeX is used is to quote a post where someone has used it to see what code(s) they have used, then use the back button on your browser to avoid submitting the post.

edit 2: Another way to view the code is to right-click an expression, then click "Show Math As", then click "TeX Commands" and a pop-up will display the code.

 

[kneaiak]

ah, for some reason I had sin2pi = to 1. Thanks for all the help. I'm sure I'll have millions of questions. Rough jumping back into this stuff after years of not seeing it. Thanks again.

 

[MarkFL]

Glad to help! You are off to a good start here...you posted the problem in its entirety and showed your work. :cool: