Integrating areas to find volumes

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AmySaunders

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The problem, from WebAssign, is copied and pasted below.

It is simple, and easy to set up. Integral from 0 to sqrt pi of 2*pi*x*sin x^2

Then I struggle through the integral by hand, because our answers are required to be in exact form, and end up with lots of crazy radicals, not part of the unit circle group. How do I solve those?

If I was allowed to use my TI84, I would put the integral in that, and come up with 15.07056343. But I am really having a hard time solving it by hand, and keeping it exact.

Let S be the solid obtained by rotating the region shown in the figure about the y-axis.a = 2
b = 1
6-3-002alt.gif

Sketch a typical approximating shell. (Do this on paper. Your instructor may ask you to turn in this graph.)
 
Please reply showing what you have done, as far as you could get. Then we can see if there's any simplification possible. Thank you! ;)
 
In fact, please state exactly what the problem is. You titled this "integrating areas to find volumes" but the integral you give looks like integrating a height to find area.
 
HallsofIvy said:
In fact, please state exactly what the problem is. You titled this "integrating areas to find volumes" but the integral you give looks like integrating a height to find area.
Click to expand...


It's a volume because that area is rotated around the y axis.
 
I integrated from 0 to sqrt pi, 2pix*2sin(x^2)dx

Pull 4pi out of the integral, integrate to get -cos(x^2)/2 at sqrt pi, minus the same thing at 0.

According to my math, the answer is 4pi, or 12.56637.

However, if I put the whole thing in the calculator, I get the answer WebAssign says is correct, 15.07056343.

Thus, my hand-solving is incorrect at some point. I'm not sure if it's the integral, or just the algebra. I really need to be accurate at hand-solving the problems.


 
You're right. Actually, integral of sin (x^2) seems to be another whole can of worms! Good thing I have an interval over which to integrate.

So... integral from 0 to sqrt pi of 2*pi*x*2*sin (x^2) ...

Pull out constants, 4pi, and integral from 0 to sqrt pi of x*sin (x^2)

u=x^2 so du=2xdx, so multiply outside of integral by 1/2 and inside of integral by 2, and take integral from 1 to sqrt pi of sin u du, which is -cos u.

-cos pi-(-cos 0)=2, then multiply it by the outside constants, 1/2*4*pi=4pi, which equals 12.566, which is what I got originally, and WebAssign says is incorrect.

I still don't know what I'm doing!:confused::confused::confused:

AND I wish I could write math with the keyboard on my laptop!
 
AmySaunders said:
The problem, from WebAssign, is copied and pasted below.

It is simple, and easy to set up. Integral from 0 to sqrt pi of 2*pi*x*sin x^2

Then I struggle through the integral by hand, because our answers are required to be in exact form, and end up with lots of crazy radicals, not part of the unit circle group. How do I solve those?

If I was allowed to use my TI84, I would put the integral in that, and come up with 15.07056343. But I am really having a hard time solving it by hand, and keeping it exact.

Let S be the solid obtained by rotating the region shown in the figure about the y-axis.a = 2
b = 1
6-3-002alt.gif

Sketch a typical approximating shell. (Do this on paper. Your instructor may ask you to turn in this graph.)
Click to expand...

Look at your problem statement carefully.

It does not state what is it that you have to calculate! It just states "sketch"!!

Assuming you need to calculate volume, using disk method your integral should be:

∫0ππ∗y2∗dx\displaystyle \displaystyle{\int_0^{\sqrt{\pi}}\pi * y^2 * dx}

∫0ππ∗[2∗sin(x2)]2∗dx\displaystyle \displaystyle{\int_0^{\sqrt{\pi}}\pi * [2*sin(x^2)]^2 * dx}

\(\displaystyle \displaystyle{4\pi\int_0^\sqrt{\pi}sin^2(x^2) dx}\)

Substitute Θ = x2 → dΘ = 2√Θ dx

2π∫0πsin2(θ)θdθ\displaystyle \displaystyle{2\pi\int_0^\pi \frac{sin^2(\theta)}{\sqrt{\theta}} d\theta}

π∫0π1−cos(2θ)θdθ\displaystyle \displaystyle{\pi\int_0^\pi \frac{1- cos(2\theta)}{\sqrt{\theta}} d\theta}

Now continue.....
 
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