cylindrical coordinatesintegrationmultivariable-calculus
I am trying to find the volume of the solid enclosed by three cylinders given by $x^2+y^2=1$, $x^2+z^2=1$, and $y^2+z^2=1$. I'm supposed to be using a triple integral, and I assume, cylindrical coordinates.
So far, I've figured out that I need to evaluate a triple integral of $dV$, which is equal to $rdzdrd\theta$. However, I am having trouble figuring out what bounds to use for $r, \theta, z$.
Any assistance or hints with this problem would be greatly appreciated!
You know the equation of such part of the sphere is $$z^2=4-(x^2+y^2),x\in[0..2],y\in[0..2]$$ But $r^2=x^2+y^2$ and then $z=\sqrt{4-r^2}$. The ranges of our new variables are : $$\theta|_0^{\pi/2}, r|_0^2, z|_0^{\sqrt{4-r^2}}$$ So we have to evaluate $$\int_0^{\pi/2}\int_0^2\int_0^{\sqrt{4-r^2}}dv$$
Answer using Cylindrical Coordinates:
Volume of the Shared region =
Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = $\sqrt{\frac{3}{4}}$.
Now the sphere is shifted by 1 in the z-direction, Hence
Volume of the Shared region = $$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} \int_{1-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
$$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$$
substitute $$u = 1-r^2 ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$
$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$
$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$ $$V =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$
$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$
Best Answer
Take the cylinders as solid, so $x^2+y^2 \le 1$ and so on.
Take for symmetry the first octant, i.e. $0 \le x,y,z$.
Then change the coordinates to cylindrical to obtain $$ \left\{ \matrix{ 0 \le r\,(by\,def.) \hfill \cr 0 \le r\cos \theta ,r\sin \theta ,z \hfill \cr r^2 \left( {\cos ^2 \theta + \sin ^2 \theta } \right) \le 1 \hfill \cr z^2 + r^2 \cos ^2 \theta \le 1 \hfill \cr z^2 + r^2 \sin ^2 \theta \le 1 \hfill \cr} \right. $$
and simplify to $$ \left\{ \matrix{ 0 \le \theta \le \pi /2 \hfill \cr 0 \le r \le 1 \hfill \cr 0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta } \hfill \cr 0 \le z \le \sqrt {1 - r^2 \sin ^2 \theta } \hfill \cr} \right. $$
To solve for the last two bounds in $z$, again using symmetry, just reduce the angle to $\pi /4$, and integrate with these conditions $$ \left\{ \matrix{ 0 \le \theta \le \pi /4 \hfill \cr 0 \le r \le 1 \hfill \cr 0 \le z \le \sqrt {1 - r^2 \cos ^2 \theta } \hfill \cr} \right. $$
Then finally you shall multiply by $16$.