# Integrals and Spherical Co-ordinates

I just do not understand how the spherical co-ordinates conversion system works. I understand the concept, but the finding the limits for p,φ,θ does not work for me (I study part-time by myself).

The question is: " Let D be the 3-Dimensional region inside the sphere $$x^2 + y^2 + z^2 = 4$$ above the cone $$z= \sqrt{4x^2 + 4y^2}$$

"Attempt at answer": Function conversion: $$\iiint p^2 \,dp \,dφ \,dθ$$

The limits y (θ): It is a full enclosed circle, thus 0 < 2π

The limits of x(p) x (p): r = 2 and therefore z = 2 z = pcosφ 2 = pcosφ 2/cosφ = p p = 2secφ

Therefore the limit is from 0 to 2secφ but a website i assessed had a different value. Is this because p = r and from the formula of the sphere it is r = 2 therefore p = 2?

The limits of z (φ): $$z = √4x^2 + 4y^2$$

$$p^2cosφ^2 = 4p^2sinφ^2cosθ^2 + 4p^2sinφ^2sinθ$$

$$tanφ^2 = 1/4$$

$$tanφ^2 = \frac{1}{\sqrt(4)}$$

However what now? Dont belive that you have a tan 1/2

Thank you!

Using spherical coordinates $$x = rcos\theta cos\phi$$ $$y = rsin\theta cos\phi$$ $$z = rsin\phi$$

The sphere becomes $$r = 2$$ and the cone becomes $$tan\phi = 2$$

The volume becomes $$\int_{\theta=0}^{2\pi}\int_{\phi=tan^{-1}2}^{\frac{\pi}{2}}\int_{r=0}^{2}r^{2}cos\phi dr d\phi d\theta$$ $$= \frac{16\pi}{3}(1 - \frac{2}{\sqrt{5}})$$

• Thanks, im struggling with the pi/2 part of the integral, why pi/2? – Shaun Weinberg Apr 13 at 17:23
• $\phi$ is the angle made with the xy plane and perpendicular to the xy plane. Since the volume is above the cone intersected with the sphere, so the angle should be from $\tan^{-1} 2$ to the z axis which is $\phi = \frac{\pi}{2}$ – KY Tang Apr 14 at 1:14

You have $$tan(\phi)=1/2$$, so $$\phi=arctan(1/2)$$ and the bounds for $$\theta$$ are also correct.

Going back to the definition of $$\phi$$, which is the angle $$\rho$$ makes between the positive z axis. We are considering the region above the cone, so the upper bound of $$\phi=arctan(1/2)$$ and the lower bound starts at the z axis, so $$\phi\in[0,arctan(1/2)]$$

The region lies inside the sphere, so $$\rho$$ cannot extend past the sphere, in other words the sphere $$x^2+y^2+z^2=4$$ describes the upper bound of $$\rho$$, which you have as 2. So, $$\rho\in[0,2]$$, now you just integrate over these bounds.

• I'm with you now, ok thank you. – Shaun Weinberg Apr 13 at 17:27