calculusintegrationmultivariable-calculusspheres
Two spheres, one of radius $1$ and one of radius $\sqrt{2}$, have centres that are $1$ unit apart. Find the volume of the smaller region that is outside one sphere and inside the other.
Can use either spherical or cylindrical coordinates. The $2$ spheres can be anywhere in three dimensional space. Apparently using the correct coordinates will lead into an easy integration. Been tossing around the two and haven't seen any of them working out to be easy.
First find where the cone intersects the inner sphere:
$$x^2+y^2=1-z^2$$
$$z^2=3-3z^2$$
$$z^2=\frac{3}{4}$$
This means that the radius of the boundary between the cone and inner sphere is $\frac{1}{2}$, and the radius of the boundary between the cone and the radius of the boundary for the outer sphere can be found to be $\frac{\sqrt{33}}{2}$, using the same process. Converting the sphere equations into cylindrical coordinates and using as bounds $0 \le \theta < 2\pi$ and $\frac{1}{2}\le r < \frac{\sqrt{33}}{2}$, the volume integral is as follows:
$$V=\int_0^{2\pi}\int_{\frac{1}{2}}^{\frac{\sqrt{33}}{2}}(\sqrt{9-r^2}-\sqrt{1-r^2})rdr d\theta$$
A similar procedure can be followed to find the integral in rectangular and spherical coordinates.
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
The region whose volume $V$ is to be found is obtained by revolving the shaded area around the $y$-axis. For $x$ between $-1$ and $1$, the upper bounding curve has equation $y=\sqrt{1-x^2}$, while the lower bounding curve has equation $y=\sqrt{2-x^2}-1$. Using shell integration we may calculate $V$: $$V=2\pi\int_0^1x(\sqrt{1-x^2}-\sqrt{2-x^2}+1)\ dx$$ $$=2\pi\int_0^1x\sqrt{1-x^2}-x\sqrt{2-x^2}+x\ dx$$ $$=2\pi\left[-\frac13(1-x^2)^{3/2}+\frac13(2-x^2)^{3/2}+\frac12x^2\right]_0^1$$ $$=2\pi\left(\frac56-\frac{2\sqrt2-1}3\right)$$ $$=\frac{(7-4\sqrt2)\pi}3$$