Double integral in polar coordinates 3

**Yan** · May 24th 2008, 04:50 PM

Use polar coordinates to find the volume of the given solid.

Bounded by the paraboloids z=3x^2 + 3y^2 and z=4-x^2-y^2

**galactus** · May 24th 2008, 05:17 PM

$3(x^{2}+y^{2})=4-(x^{2}+y^{2})$

$3r^{2}=4-r^{2}$

$r=\pm{1}$

$\int_{0}^{2\pi}\int_{0}^{1}\int_{3r^{2}}^{4-r^{2}}rdzdrd{\theta}$

Here is the same in rectangular. That way you can see how they interchange.

$3x^{2}+3y^{2}=4-x^{2}-y^{2}$

Solve for $y=\pm\sqrt{1-x^{2}}$

From this we can see $x=\pm{1}$

So, we get $\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\int_{3x^{2}+3y^{2}}^{4-x^{2}-y^{2}}dzdydx$

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