1. The problem statement, all variables and given/known data 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. 2. Relevant equations 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). 3. 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!