Surface Area Using Integrals

Question

Question: Find the surface area of the portion of the sphere x^2 + y^2 +z^2 = 4 that lies inside the cylinder x^2 + y^2 = 2y

My work: 

Let f(x,y) = sqrt(4- x^2 - y^2)

F(x,y,z) = f(x,y) - z 

Required surface area can be calculated using 

S = integrate over theta and r (2/(sqrt(4 - r^2))) rdr d(theta), where limit of r, theta as follows:

r: 0 - rsin(theta)

theta: 0 - 2pi

Is this correct?

Answer 1

At the outset, it may not be clear which coordinate system leads to an easier solution.

But what gives us a clue is that we are trying to calculate the area of a part of a sphere.

So constructing a differential surface in spherical coordinates seems like the natural choice, which leads to smoother calculations, (assuming that you know how to transform Cartesian coordinates to spherical coordinates).

Transform equations to spherical coordinates

The given sphere equation

x² + y² + z² = 4

in spherical coordinates (r, θ, φ) becomes

r² = 4 eq. 1

(We transform the cylinder equation first to cylindrical coordinates and then to spherical coordinates, to make the process easier.)

And the given cylinder equation

x² + y² = 2y

in cylindrical coordinates (ρ, φ, z) becomes

ρ² = 2 ρ sin φ

because y = ρ sin φ in cylindrical coordinates.

Now transforming the last equation to spherical coordinates, we get

r sin θ = 2 sin φ eq. 2

because ρ = r sin θ.

Find the intersecting curve (and the integration limits)

Now we can find the intersecting curve inside of which lies the surface, the area of which we would like to calculate.

The intersecting curve of the sphere and the cylinder yields from solving the system of eq. 1 and eq. 2:

r² = 4

r sin θ = 2 sin φ

From the first equation, we have r = 2 (remember that r is always positive in spherical coordinates, r ≥ 0 )

Substituting the value of r = 2 in the second equation, we get

sin θ = sin φ eq. 3

which is the equation of the intersecting curve.

eq. 3 implies that 0 ≤ φ ≤ π because in spherical coordinate system 0 ≤ θ ≤ π. That sets our integral limits when calculating the surface area.

Construct differential surface area

To calculate the differential surface area dS in spherical coordinates (r, θ, φ), we notice that the radius r remains constant, so only θ and φ change. That is, dS only involves dθ and dφ. But how exactly?

Well, moving in the direction of dθ, the differential curve length will be

r dθ. exp. 4

And moving in the direction of dφ, the differential curve length will be ρ dφ which is the same as

r sin θ dφ exp. 5

because ρ = r sin θ.

exp. 4 and exp. 5 can be derived by drawing the differential surface (or the differential volume, in the general case) in a coordinate system.

Multiplying exp. 4 and exp. 5 with each other, the differential surface area becomes

dS = (r dθ)(r sin θ dφ)

dS = r² sin θ dθ dφ

dS = 4 sin θ dθ dφ eq. 6

Remember that r = 2.

Integrate dS to calculate S

Now we integrate dS over the limits that we had previously derived to calculate S.

S = ∫ dS

S = ∫_0→π ∫_0→π 4 sin θ dθ dφ

S = 4 ∫_0→π ( ∫_0→π sin θ dθ ) dφ

where the inner integral becomes

∫_0→π sin θ dθ = - cos π - (- cos 0) = -(-1) - (-1) = 2

Hence,

S = 4 ∫_0→π 2 dφ

S = 8 ∫_0→π dφ

S = 8 (π - 0)

S = 8π

which is the final answer.

P.S.

Notice that this is the area of two sections of the sphere that cap the cylinder. One at the top, and the other at the bottom.

The two caps are symmetrical with respect to the x-y plane.

If we were asked to calculate the surface area of one the caps, the answer would have been half of the calculated amount, that is 4π.