Calculating Volume of a Double-Lobed Cam Using Polar Coordinates

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SUMMARY

The volume of a double-lobed cam modeled by the inequalities \(\frac{1}{4} \leq r \leq \frac{1}{2}(1 + \cos 2\theta)\) and \(-\frac{9}{4(x^2 + y^2 + 9)} \leq z \leq \frac{9}{4(x^2 + y^2 + 9)}\) can be calculated using polar coordinates. The integration involves evaluating the double integral \(\iint_R (z_{upper} - z_{lower}) \, dA\) with bounds for \(r\) from \(\frac{1}{4}\) to \(\frac{1}{2}(1 + \cos 2\theta)\) and \(\theta\) from \(0\) to \(2\pi\). The calculated volume is approximately 0.79993, which is reasonable given the dimensions of the cam.

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Homework Statement


The surface of a double lobed cam are modeled by the inequalities:

\frac{1}{4}\leqr\leq\frac{1}{2}(1+cos2θ)

and

-9/(4(x2+y2+9)) ≤ z ≤ 9/(4(x2+y2+9))

Find the volume of the steel in the cam.


Homework Equations





The Attempt at a Solution


I know I need to use a double or triple integral to solve this. I was thinking since I was given r I could change to polar coordinates and solve that way.

Please give me some guidance.
 
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Hint: General formula for volumes:
$$\iint_R z_{upper}-z_{lower}~dA$$
 
Ok, that makes sense. Wasn't sure if I could do that.

R would be the r given. Theta is from 0 to 2*pi.

Therefore I can convert the bounds to polar coordinates.

My z_upper - z_lower is taken from the z given in the inequality.

x^2 + y^2 become r^2 and I can integrate completely from there.

Right?
 
If you mean what I think you mean, yes.
 
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Ha ha ok. Thanks LCKurtz.

When I evaluate the integral (using a calculator) I get 0.79993.

This seems awfully low to be a volume of an shape like this.

- I checked it twice for errors, I think it's accurate.
 
That looks like it might be about right. Here's a picture (I had a little time to waste):
attachment.php?attachmentid=68156&stc=1&d=1396220575.jpg
 

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Wow, thanks for wasting your time for me! ;-)

I guess the cam is pretty small so the volume is more reasonable than I thought.
 

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