- #1
mreaume
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Homework Statement
Find the area of that part of the cylinder x^2 + y^2 = 2ay that lies inside the sphere x^2 + y^2 + z^2 = 4a^2.
Homework Equations
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If a surface S can be parametrized in terms of two variables u and v, then dS = Norm[dR(u,v)/du x dR(u,v)/dv].
The surface area is given by ∫∫dS where the integral is evaluated over the given region (here it would be over the region of the cylinder that lies within the sphere).
The Attempt at a Solution
I completed the square of the cylinder to get x^2 + (y-a)^2 = a^2
I parametrized the cylinder as follows (where t stands for theta):
R(t,z) = (rcos(t),rsin(t),z)
r = 2asin(t)
So R(t,z) = (2asintcost,2asintsint,z)
I computed dS = Norm[dR(t,z)/dt x dR(t,z)/dz] and got 2adtdz, which I know is correct (same answer as textbook).
For the given cylinder, t goes from 0 to Pi. By equating x^2 + y^2 in the equations given (i.e. sphere and cylinder), we get:
2ay = 4a^2 - z^2
z^2 = 4a^2 - 2ay
z^2 = 4a^2 - 2a(2a(sint)^2)
z^2 = 4a^2(1-(sint))^2
z^2 = 4a^2((cost)^2)
z = 2acost
So z goes from -2acost to 2acost.
So the Area = ∫∫1dS
= ∫∫2adzdt, where t goes from 0 to Pi and z from -2acost to 2acost.
We get:
A = ∫2a(2acost + 2acost)dt
A = ∫2a(4acost)dt
A = ∫8a^2costdt
A = 8a^2∫cost
But the integral of cost from 0 to pi = 0.
So A = 0.
The textbook uses symmetry and integrates from 0 to pi/2 (i.e. finds the area of half of the cylinder under the sphere), and then multiplies by 2. Which makes sense. But why doesn't this method work? Furthermore, ∫cost from 0 to pi is NOT EQUAL to 2∫cost from 0 to pi/2, so the two methods are not compatible. So what did I do wrong?
Thanks!