# Homework Help: Surface Integral Problem

1. Dec 3, 2013

### cp255

If the mass per unit area of a surface is given by ρ=xy, find the mass if S is the part of the cylinder x2+z2=25 which is in the first octant and contained within the cylinder x2+y2=16.

So here was my attempt.

I parametrized the curve.
x2+z2=25
r(u, v) = <5cos(u), v, 5sin(u)>

I then plugged into the bounds
x2+y2=16.
25cos2(u) + v2 = 16
v = sqrt(16 - 25cos2(u))

Next I took the cross product of ru X rv. Its magnitude is a constant 5.

Now I solved the integral with bounds 0 < u < Pi/2 and 0 < v < sqrt(16 - 25cos2(u))
∫∫25 v cos(u) du dv

This returns the result of -25/3 but since we are looking for a mass I submitted the answer of positive 25/3 and this is wrong. I checked the integration on my calculator and it gets the same result.

2. Dec 3, 2013

### LCKurtz

$u$ doesn't go from $0$ to $\pi/2$. Look at a picture showing what angle $u$ represents.

3. Dec 4, 2013

### cp255

I redid my bounds and now I have arccos(sqrt(16-v^2)/5) < u < pi/2 and 0 < v < 4 which results in the answer of 22/15 which is still wrong. I fell like I found the wrong value for u.

4. Dec 4, 2013

### LCKurtz

I get 22/3 both that way and working the integral in reverse order so I think you have a mistake somewhere. But haven't you left out multiplying by the multiplier $|r_u\times r_v|$? That would give 110/3 by my calculations. Is that the given answer?