Find the surface area of the part of the sphere $x^2 + y^2 + z^2 = 16$
inside the cylinder $x^2 – 4x + y^2 = 0$
First of all, I need to find the equation of the plane along which these two solids intersect. Solving for $y^2$ from the equation of the cylinder and substituting to the sphere, we get
$$z^2+4x=16$$
And so we have the integration are with respect to $z$ and $x$. The transform we want are considering is
$$T(x,z) = (x, z, 16-x^2-z^2)$$
We are integrating on the $z^2+4x=16$ surface. and so we need to set bounds for $z$ and $x$.
$$z_0 = 0 \le z \le z_1 =4 $$
$$x_0 = -\sqrt{16-z^2} \le x \le x_1 = \sqrt{16-z^2}$$
Because of symmetry, we can find the are above the $xy$ plane and then multiply by $2$.
And so, using the formula for the surface integral, we get that the surface are
$$S = \int_{z_0}^{z_1}\int_{x_o}^{x_1}\sqrt{1 + (\frac{\partial y}{\partial z})^2+ (\frac{\partial y}{\partial x})^2}$$
Is my attempt to solve this correct? I have never calculated a surface integral before and so I am not sure about the bounds and the method.
Best Answer
I'm not sure what exactly you are doing here. You don't want/need the curve of intersection. You want to set up a surface integral over the sphere, but you need to set up the bounds such that you only integrate over that part of the sphere which lies within the cylinder.
The cylinder has an easy representation in polar coordinates (verify that it's $r=4\cos t$) and the projection of the part of the sphere you want onto the $xy$-plane is therefore the disc $D$ given by the bounds $0 \le r \le 4\cos t$ and $-\tfrac{\pi}{2} \le t \le \frac{\pi}{2}$ in polar coordinates.
Converting the upper half of the sphere into these coordinates gives $z=\sqrt{16-r^2}$ so you can parametrize that upper half of the sphere as: $$\vec \phi(r,t) =\left(r,t,\sqrt{16-r^2}\right)$$ For the surface integral, you compute: $$\left\| \vec \phi_r \times \vec \phi_t \right\| = \ldots = \frac{4}{\sqrt{16-r^2}}$$ So you get (doubling for the bottom half): $$\mbox{Surface area}=2\iint_S \mbox{d}S=2\iint_D \left\| \vec \phi_r \times \vec \phi_t \right\| \mbox{d}A = 2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^{4\cos t}\frac{4}{\sqrt{16-r^2}} r\,\mbox{d}r\,\mbox{d}t$$