Find the volume of the solid in the first octant bounded by the coordinate planes, the cylinder $x^2 +y^2 =4$ and the plane $z+y=3$.

I found the integral bounds just fine. So I have

$\int_{0}^{2} \int_{0}^{\sqrt{4-y^2}}\int_{0}^{3-y}dzdxdy$

$ = \int_{0}^{2} \int_{0}^{\sqrt{4-y^2}} (3-y) dxdy$

Now I saw a method used here where we let $ y = rsin\theta$ and $x = rcos\theta$ (I think)

And this lead to the following expression:

$ = \int_{0}^{\frac{\pi}{2}} \int_{0}^{2} (3-rsin\theta)r \space drd\theta$

I'm not sure where the second $r$ came out of, I'm also unsure of how the bounds changed the way they did. Why don't we sub $r = 2$ for all $r$ in the integral?

Could someone explain to me how the above integral is true by showing a few extra steps in between.

Thanks for the help.


The rectangular coordinates $x$ and $y$ have been changed to polar coordinates. Look this up in your class notes or the text you got this from. The "second $r$" comes out of that change of coordinates as well as the limits of integration. It is best to draw a picture of the situation in the $xy$-plane for this. See also http://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_.28area.29

  • $\begingroup$ That does sound familiar. I'll look it up, thanks. $\endgroup$ May 4 '15 at 14:41

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