3
$\begingroup$

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.

  • 2
    [Here, have a nice picture...](http://i.stack.imgur.com/tFs6w.png)2012-05-13
  • 0
    @J.M. Thanks. Do you mind sharing the code?2012-05-13
  • 0
    I used the `RegionPlot3D[]` function in *Mathematica* and gave it your set of inequalities...2012-05-13

2 Answers 2

2

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: $$ \mathrm{Vol}(D)=2\mathrm{Vol}(D_1)=2\iint\limits_{x^2+y^2\leq 4, x \geq 0, y \geq 0}ydxdy= 2\int_0^2\int_0^{\frac{\pi}{2}}\rho\sin\varphi\rho d\varphi d\rho=\frac{16}{3} $$

  • 0
    I believe that $dρdφ$ should be inverted and the result 8/3.2012-05-13
  • 0
    @sadma12 Please check your calculations. Both integrals - mine and Ayman Hourieh's gives 16/32012-05-13
  • 0
    I didnt consider the 2 in front of the integral. It is 16/3. My mistake.2012-05-13
2

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}]
  • 0
    It seems you're using *Mathematica*. You might want to consider the `RegionPlot3D[]` function...2012-05-13
  • 0
    @J.M. Thanks for the tip. I didn't know about this function. I've updated my answer.2012-05-13