Triple integral doubt, spherical coordinates

[Telemachus]

Hi there. I have some doubts about this exercise. It asks me to use the appropriated coordinates to find the volume of the region indicated below. The region is determined by the first octant under the sphere $$x^2+ y^2+ z^2=16$$ and inside the cylinder $$x^2 +y^2=4x$$
The first thing I did was to complete the square with the cylinder: $$(x-2)^2 +y^2=4$$
And then I thought of using spherical coordinates, this way:

$$\displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\displaystyle\int_{0}^{\displaystyle\frac{\pi}{4}}\displaystyle\int_{0}^{\sec(\theta)}\rho^2 \sin \phi d\rho d\theta d\phi + \displaystyle\int_{0}^{\displaystyle\frac{\pi}{2}}\displaystyle\int_{0}^{\displaystyle\frac{\pi}{4}}\displaystyle\int_{\sec(\theta)}^{4\cos\theta}\rho^2 \sin \phi d\rho d\theta d\phi$$

Is this okay?

Bye there, thanks for anyhelp.

Edit: mm, now I see it isn't. Unless I think its not, because it doesn't defines the region under the sphere. I think I should use cylindrical coordinates.

 

[lippyka]

That's right, you should use cylindrical coordinates. The region is bounded by the sphere x^2 + y^2 + z^2 = 16 and the cylinder x^2 + y^2 = 4x, so you need to integrate over the region defined by 0 ≤ ρ ≤ 4cosθ and 0 ≤ θ ≤ π/2 and 0 ≤ z ≤ √(16 − ρ2). To do this, you can use the following integral:\int_0^{\frac{\pi}{2}}\int_0^{4\cos{\theta}}\int_0^{\sqrt{16-\rho^2}} \rho dz d\rho d\theta