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I have this weird integral to find. I am actually trying to find the volume that is described by these two equations.

$x^2+y^2=4$ and

$x^2+z^2=4$ for

$x\geq0, y\geq0, z\geq0$

It is a weird object that has the plane $z=y$ as a divider for the two cylinders. My problems is that I can't find the integration limits.

I can't even draw this thing properly.

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    I used the `RegionPlot3D[]` function in *Mathematica* and gave it your set of inequalities...2012-05-13

2 Answers 2

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From the symmetry it is enough to find volume of the half $D_1$ of this domain $D$. This half is described by inequalities $ D_1:x^2+y^2\leq 4,\quad x \geq 0,\quad y \geq 0,\quad z\leq y. $ To find its volume use polar coordinates: $ \mathrm{Vol}(D)=2\mathrm{Vol}(D_1)=2\iint\limits_{x^2+y^2\leq 4, x \geq 0, y \geq 0}ydxdy= 2\int_0^2\int_0^{\frac{\pi}{2}}\rho\sin\varphi\rho d\varphi d\rho=\frac{16}{3} $

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    I didnt consider the 2 in front of the integral. It is 16/3. My mistake.2012-05-13
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Since $z \ge 0$, we can rewrite $x^2 + z^2 = 4$ as: $ z = \sqrt{4 - x^2} $

This is the function of integration, and the area is:

$ \mathcal{A} = \{(x, y): x^2 + y^2 \le 4, x \ge 0, y \ge 0\} $

The volume is:

$ V = \iint_{\mathcal{A}} \sqrt{4 - x^2} \, dx dy $

Can you calculate this integral via polar coordinates?

Here is a plot of the boundaries of the object:

boundaries

And here is the object itself: (Thanks J.M. for the Mathematica tip) object

Mathematica code:

RegionPlot3D[x^2+y^2<=4&&x^2+z^2<=4,{x,0,2},{y,0,2},{z,0,2}] 
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    @J.M. Thanks for the tip. I didn't know about this function. I've updated my answer.2012-05-13