Surface area of sphere $x^2 + y^2+ z^2 = 16z$ within paraboloid $z = x^2 + y^2$

Question

How would you calculate the surface area of the portion of the sphere $x^2 + y^2+ z^2 = 16z$ that lies within the paraboloid $z = x^2 + y^2$.

Points common to the sphere and paraboloid satisfy the equation $z + z^2 = 16z$, so there we have either z = 0 or z = 15. The former corresponds to the origin (0, 0, 0) which lies in both surfaces where they both have z = 0 as a tangent plane.

However I am not sure how to continue. Could anyone please guide me through how you would attempt this question? thank you

Answer 1

Simplify the task by re-centering the circle with $w=z-8$. Then, we have the two equations,

$$x^2+y^2+w^2=64,\>\>\>\>\>x^2+y^2=w+8$$

which intersect at

$$w^2+w-56=0\implies w = 7,\>\>\>x^2+y^2=15$$

The surface integral is,

$$I = \int_{A} \sqrt{1+(w_x')^2+(w_y')^2}dA=\int_{A} \frac 8wdA$$

Evaluate the integral in polar coordinates,

$$I = 8\int_0^{2\pi} d\theta\int_0^{\sqrt{15}} \frac{rdr}{\sqrt{64-r^2}}=-16\pi\sqrt{64-r^2}|_0^\sqrt{15}=16\pi$$

Answer 2

Both the sphere and the paraboloid are rotationally symmetric around the $z$-axis, so their intersection must be one or more circles (again symmetric around the $z$-axis). You can now determine a parametrization of the part of the sphere between those circles and integrate the surface element.

For example you can say $s(u,v) = (c_x + r\cos u\cos v, c_y + r\sin u\cos v, c_z + r\sin v)$ for proper values of radius $r$ and center $c = (c_x,c_y,c_z)$, then choose the domains of $u$ and $v$ as necessary to cover exactly the parts of the sphere you need.

Answer 3

This figure should help:

Calculate the angle $\phi$ with respect to the $z$ axis of the (red) boundary, then you can easily calculate the area of the (outer) sphere.

Knowing its $z$ height allows you to calculate the area of the paraboloid too.