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Compute the volume of solid enclosed between the surfaces $x^2+y^2=9$ and $x^2+z^2=9$

What should be the limits of the integrals ?

I am getting this

z from $-\sqrt{9-x^2}$to $\sqrt{9-x^2}$

x from $-\sqrt{9-y^2}$ to $ \sqrt{9-y^2}$

y from $0$ to $3$

Is this correct?

Aman Mittal
  • 2,091

2 Answers2

6

As you noted, you are looking for a triple integration, but I am posting another parallel approach. If we consider the whole volume, we have $8$ times of the following solid thing(right on the top). This means that, I consider the first $1/8$ of $\mathbb R^3$ for working. And, I am assuming you can find out why the thin slab is really an square (Just examine the equations of both Cylinder). Moreover, we can move that piece from $x=0..x=3$ and so:

$$V_{1/8}=\int_0^3S_{\small{\text{slab}}}(x)dx=\int_0^3(9-x^2)dx$$

I think this way is easier than doing triple integrals, however, you may not like it. So just keep it for the later and let's wait for your desire solution. :-)

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Mikasa
  • 67,374
1

The volume of any region $W$ can be found using triple integrals as follows:

$$V=\iiint_WdV$$

Your boundaries are almost correct. However you are limiting $y$ to the positive $y$ axis, which is only a half-sphere. The interval for $y$ should be $-3\le{y}\le{3}$.