Setting up triple integral

**PMatt** · December 9th 2009, 03:49 PM

Edit: Title should read double integral... oops

Having a little trouble figuring this one out.

Region bounded by 4x^2 + y^2 = 4z and z = 2

Thanks.

**Scott H** · December 9th 2009, 04:14 PM

The two surfaces intersect when

$4x^2+y^2=4\cdot 2=8.$

What shape does this define? Would it be easier to use Cartesian or polar coordinates?

**PMatt** · December 9th 2009, 07:31 PM

I was assuming that I should use polar coordinates, but I'm still not quite sure how to set it up.

**Scott H** · December 11th 2009, 03:45 AM

The equation that bounds the region of integration can be rewritten

$\left(\frac{x}{\sqrt{2}}\right)^2+\left(\frac{y}{2 \sqrt{2}}\right)^2=1.$

This defines an ellipse whose equation in polar coordinates is

$r=\sqrt{2}\cos\theta + 2\sqrt{2}\sin\theta.$

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