Calculating Mass for a Cylinder in the First Octant

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SUMMARY

The discussion focuses on calculating the mass of a surface defined by the cylinder equation x² + z² = 25 in the first octant, constrained by the cylinder x² + y² = 16, with a mass per unit area given by ρ = xy. The user initially parametrized the surface and set up the integral but encountered discrepancies in the results. After adjusting the bounds for u and v, the user arrived at an answer of 22/15, which was still incorrect. The discussion highlights the importance of including the magnitude of the cross product |ru × rv| in the final calculation, which led to a proposed answer of 110/3.

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cp255
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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.
 
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cp255 said:
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.

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

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