2

To be honest I'm not even too sure of what I'm integrating. I'm picturing two spheres touching each other, with a cylinder of two different radii going from the center of one to the other and I'm supposed to calculate the volume of the space inside the cylinder not occupied by the spheres. Is that correct?

enter image description here

I'm still stuck though, I don't really know how to start this.

EDIT - Accidentally wrote the wrong integral.

xsr
  • 167
  • 2
    The region you're considering is a spherical shell: it is outside a sphere centered at the origin with radius 2 and inside a sphere centered at the origin with radius 3. – user84413 Jul 16 '15 at 00:30
  • 1
    They don't touch each other; rather they are concentric, i.e. they both have the same center. One is inside the other. ${}\qquad{}$ – Michael Hardy Jul 16 '15 at 00:43
  • 1
    Please, do not change your question after a valid answer has been provided. If you discover a mistake in your problem statement, then pose a new problem. – Ron Gordon Jul 16 '15 at 06:03

2 Answers2

2

The spheres are concentric, so the integral is straightforward in spherical coordinates:

$$\int_2^3 dr \, r^6 \, \int_0^{\pi} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi \, = \frac{4 \pi}{7} (3^7-2^7) = \frac{8236 \pi}{7}$$

Ron Gordon
  • 138,521
  • NB This solution was posed before the OP edited the question. I leave this as the answer to the original question. – Ron Gordon Jul 16 '15 at 06:03
1

It is well-known that the Cartesian coordinates may be retrieved from the spherical coordinates (radius r, inclination $\theta$, azimuth $\varphi$), where $r \in [0, \infty)$, $\theta \in [0, \pi]$, $\varphi \in [0, 2\pi)$, by: $$x=r \, \sin\theta \, \cos\varphi \\ y=r \, \sin\theta \, \sin\varphi \\ z=r \, \cos\theta$$ The spheres are concentric, then $2\leq r\leq 3$, while $$x^2+y^2=\left(r \sin\theta\cos\varphi\right)^2+\left(r \sin\theta\sin\varphi\right)^2=r^2\sin^2\theta$$ and $$\mathrm{d}V=r^2 \sin \theta \,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\varphi.$$ Then $$\displaystyle\iiint_E (x²+y²) \;\mathrm{d}V=\int_2^3 \int_0^\pi\int_0^{2\pi}r^2\sin^2\theta \cdot r^2 \sin \theta d\varphi d\theta dr$$

Mark
  • 7,841
  • 6
  • 38
  • 72