find the volume of the paraboloid z=x^2+y^2 below the plane z=4.

i am stuck on setting up the integral..can any one help???

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- Jul 6th 2008, 07:45 PMchris25volume of paraboloid
find the volume of the paraboloid z=x^2+y^2 below the plane z=4.

i am stuck on setting up the integral..can any one help??? - Jul 6th 2008, 09:14 PMChris L T521
Let the cross section of the paraboloid at z=4 be our region R.

Thus, the x and y limits of integration become pretty obvious: $\displaystyle \left\{R: \ -2\leq x \leq 2, \ -2\leq y \leq 2\right\}$.

http://img.photobucket.com/albums/v4...LT521/Circ.jpg

(this is the region R)

Thus, our volume integral would be $\displaystyle \int_{-2}^2\int_{-2}^2 x^2+y^2 \,dx\,dy$.

We can also represent this integral in polar coordinates. Note that the limits of integration change!

This time we see : $\displaystyle \left\{R: \ 0\leq r \leq 2, \ 0\leq \theta \leq 2\pi\right\}$.

As a result, the integral becomes $\displaystyle \int_0^{2\pi}\int_0^2 r^3 \,dr\,d\theta$.

Hope that this makes sense!

--Chris - Jul 6th 2008, 11:42 PMJhevon
Hey, Chris answered Chris! :Dactually, wouldn't the volume be given by $\displaystyle V = \int _{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \int_{x^2 + y^2}^4 ~dzdydx= \int _{-2}^2 \int_{- \sqrt{4 - x^2}}^{\sqrt{4 - x^2}} \bigg[4 - (x^2 + y^2) \bigg]~dydx$

which yields $\displaystyle \int_0^{2 \pi} \int_0^2 (4r - r^3)~dr d \theta$ ?

...or am i (Drunk) again?