Any help would be great (esp. with setting up bounds and the actual integral)
A sphere of radius 3 passes through the center of a sphere of radius 1. Find the volume of the region inside the sphere of radius 3 and outside the sphere of radius 1.
Any help would be great (esp. with setting up bounds and the actual integral)
A sphere of radius 3 passes through the center of a sphere of radius 1. Find the volume of the region inside the sphere of radius 3 and outside the sphere of radius 1.
Let Sphere#1 be centered at (0,0,0) with radius = 1. Let Sphere#2 be centered at (0,0,3) with radius = 3. Clearly Sphere#2 intersects Sphere#1 through its center. The equations for these sphere are described respectively: $\displaystyle x^2+y^2+z^2 = 1 \mbox{ and }x^2+y^2+(z-3)^2 = 9$. By drawing a picture you should see that when these two spheres intersect the smaller sphere is the top surface and the larger sphere is the lower surface. To find the volume of their intersection we need to know the upper and lower surfaces (and we just did that) and the region over we are intergrating. If you open parantheses in the second equation and substitute the first you get $\displaystyle -6z = - 1$ thus $\displaystyle z=1/6$ this tells us the intersection occurs at the constant height of $\displaystyle 1/6$. Thus, the region of integration is $\displaystyle x^2+y^2 + \frac{1}{36} = 1\implies x^2 + y^2 = \frac{35}{36}$.
Hello, ratedmichael!
[size=3]A sphere of radius 3 passes through the center of a sphere of radius 1.
Find the volume of the region inside the sphere of radius 3 and outside the sphere of radius 1.We have a cross-section of the spheres.Code:3| * * * * | * * | : * * | : *o | :o o * | o * o - - * - - - - - + - - - o - * - o - - -3* | 2o *3 o | :o o * | : *o * | : * * | * * * * |
We will revolve the appropriate regions about the x-axis.
The large circle has equation: .$\displaystyle x^2+y^2 \:=\:9\quad\Rightarrow\quad y_1^2 \:=\:9-x^2$
The small circle has equation: .$\displaystyle (x-3)^2 + y^2 \:=\:1\quad\Rightarrow\quad y_2^2 \:=\:-x^2+6x-8$
On $\displaystyle [-3,\,2]$, we revolve $\displaystyle y_1$ about the x-axis:
. . $\displaystyle V_1 \;=\;\pi\int^2_{-3}(9 - x^2)\,dx$
The circle intersect at $\displaystyle x \:=\:\frac{17}{6}$
On $\displaystyle \left[2,\,\frac{17}{6}\right]$, we find the difference of the volumes.
. . $\displaystyle V_2\;=\;\pi\int^{\frac{17}{6}}_2\left[y_1^2 - y_2^2\right]\,dx \;=\; \pi\int^{\frac{17}{6}}_2\bigg[(9-x^2) - (-x^2 + 6x - 8)\bigg]\,dx$
Got it?
I may be wrong, but 6.545 seems a little small to be the volume inside the larger sphere and outside the smaller one. It is 27 times the volume of the smaller. The volume of the big sphere is 36Pi = 113.1. Even with the overlap of the smaller sphere, one would think there is more than 6.545 cubic units in it.
There is a formula for the volume of two intersecting spheres. I seen it somewhere so I wrote it down.
We have two spheres of radius R=3 and r=1
With equations $\displaystyle x^{2}+y^{2}+z^{2}=9$
$\displaystyle (x-a)^{2}+y^{2}+z^{2}=1$
In this case, $\displaystyle (x-3)^{2}+y^{2}+z^{2}=1$
The formula is thus:
$\displaystyle V=\frac{{\pi}(R+r-a)^{2}(a^{2}+2ar-3r^{2}+2aR+6rR-3R^{2})}{12a}$
Entering in our R=3, r=1, a=3, we get $\displaystyle V=\frac{7{\pi}}{12}$
This is the volume inside the intersection of the spheres.
Subtract it from the volume of the larger to get the volume inside the larger but outside the smaller.
$\displaystyle 36{\pi}-\frac{7{\pi}}{12}=\boxed{\frac{425{\pi}}{12}\appro x{111.265}}$