Volume of Region Paraboloids

Question

How do I find the volume of the solid region which is bounded by $z=2x^2+2y^2$ and $z=3-x^2-y^2$?

So I first realized that these two functions are paraboloids and I have to find the volume of their intersection. But, I'm not quite sure how to do so. Can someone please help me out through each step explaining me so that I understand?

UPDATE:

So I was trying it out by myself and I set up the volume integral as:

$$\int^{2\pi}_0\int^1_0\int^{3-r^2}_{2r^2}{rdzdrd\theta}$$

Am I right?

I know that I have to find the intersection first, which is when $x^2+y^2=1$ if I'm not mistaken — aña, Jul 23 '15 at 13:01

Seth · Answer 1 · 2015-07-23T13:10:50.327

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Hint: use cylindrical coordinates. You know that the volume of the region E bounded by the paraboloids is $\iiint_E \text{d}V$. Integrate with respect to $z$ first. You will need to find the circle of intersection of the two paraboloids to set up the limits of integration of $r$ and $\theta$.

edited Jul 23 '15 at 13:10

answered Jul 23 '15 at 13:06

Seth

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Can you see my update above? – aña Jul 23 '15 at 13:07
Yup, that looks right! – Seth Jul 23 '15 at 13:07
Now can you help me integrate that. That gives me the biggest problem. – aña Jul 23 '15 at 13:08
Well start by finding the antiderivative of $r$ with respect to $z$ ($r$ is constant here). Then you need to evaluate at the limits of integration before you do the next 2 integrals. – Seth Jul 23 '15 at 13:10
Does the final answer come out to $\frac{6\pi}{4}$. Can you just verify that for me please? – aña Jul 23 '15 at 13:14

Tucker · Accepted Answer · 2015-07-23T13:16:13.097

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The two functions you describe are surfaces, their intersection will be a curve. This curve is a one dimensional set embedded in $\mathbb{R}^{3}$. What you mean by volume is 3-dimensional extent, a curve is a one dimensional object and so it does not have 3-dimensional extent. The shadow of the curve of intersection in the xy plane is the unit circle. You will need to know which surface is above the other in this region. It is $z=3-x^2-y^2$, from here you just need to integrate

$$\int\int (3-x^2-y^2-(2x^2+2y^2))dxdy$$

Because the region of integration is a circle, more natural coordinates would be polar.

$$\int_{0}^{2\pi}\int_{0}^{1}(3-3r^2)rdrd\theta=\frac{3\pi}{2}$$

edited Jul 23 '15 at 13:16

answered Jul 23 '15 at 13:08

Tucker

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Wouldn't it be $3r-3r^3$? – aña Jul 23 '15 at 13:15
I haven't heard the terminology "magnification factor" before, if you mean that the area element in polar is $dA=rdrd\theta$, then yes an $r$ should be included on $drd\theta$ which I have done. – Tucker Jul 23 '15 at 13:24
$(3-3r^2)r=3r-3r^3$ – Tucker Jul 23 '15 at 13:25

Volume of Region Paraboloids

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