Evaluate the surface Integral of $H(x,y,z) = yz$ over the part of the sphere $x^2+ y^2+ z^2 =4$ that lies above the cone $z = \sqrt{x^2 + y^2}$
My question is How can I setup the integral to calculate the surface integral for this one . I know for the conic part I can use the parametric coordinates
as:
$x =r\cos\theta$, $y =r\sin\theta$, $z =r$, But I am unable to setup the integral correctly.
Can anyone please tell me how should I tackle this problem ?
Thank you.
On
Okay so it has been a while for me but I will give it a try.
We want to evaluate a surface integral using coordinate transformation:
So let's start with the transformation:
$$\phi(u,v)= r\cdot\begin{bmatrix} cos(u)cos(v) \\sin(u)cos(v) \\ sin(v) \end{bmatrix}$$
We can calculate our normal vector: $$n^* = \phi(u,v)_u \times \phi(u,v)_v = cos(v)\cdot r^2 \cdot \phi(u,v)$$
Because $n^*$ points outwards, we can set $n=n^*$: $$\int_B f|do| = \int_G f(\phi(u,v)) \cdot ||n(u,v)||_2 d(u,v)$$
With $G=[0,\pi]\times[0,\pi]$, you should be able to solve this integral.
(I am not 100% sure about G)
edit: I did not fully read your question and G seems wrong. Hope this still helps somehow. You can find $G$ by using your additional equations.
Given the symmetry, it is convenient to set up the integral in spherical coordinates. Use $r=2$, $y=r\sin\phi\sin\theta$, $z=r\cos\theta$ and the surface element $ds = r^2\sin\theta d\theta d\phi$
$$I=\int_S H(x,y,z) ds =\int_S yz ds =r^4 \int_0^{2\pi} \int_0^{\pi/4}\sin\phi \cos\theta \sin^2\theta d\theta d\phi $$
Note that the result is zero due to the integral over $\phi$, which is not surprising because of $y$ in the integrand.