# Weird integral with cylinders

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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Here, have a nice picture... – J. M. May 13 '12 at 11:32
@J.M. Thanks. Do you mind sharing the code? – sadma12 May 13 '12 at 11:35
I used the RegionPlot3D[] function in Mathematica and gave it your set of inequalities... – J. M. May 13 '12 at 11:44

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:

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

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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It seems you're using Mathematica. You might want to consider the RegionPlot3D[] function... – J. M. May 13 '12 at 11:45
@J.M. Thanks for the tip. I didn't know about this function. I've updated my answer. – Ayman Hourieh May 13 '12 at 11:54

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 believe that $dρdφ$ should be inverted and the result 8/3. – sadma12 May 13 '12 at 12:01
@sadma12 Please check your calculations. Both integrals - mine and Ayman Hourieh's gives 16/3 – Norbert May 13 '12 at 12:08
I didnt consider the 2 in front of the integral. It is 16/3. My mistake. – sadma12 May 13 '12 at 12:12