Surface Area Problem

  • Thread starter bodensee9
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I am wondering if someone could help me with the following? I am supposed to find the surface area of the part of the sphere x^2 +y^2+z^2 that lies inside the cylinder x^2+y^2 = ax.

If I wanted to write a parametric equation for the sphere, I would use x = ρsinφcosθ and y = ρsinφcosθ and z = ρcosθ. Let r be the vector function given by xi +yj +zk. So, since the surface area = ∫∫|(dr/dφ) x (dr/dθ)|dφdθ. So if I do that with the following, I have |(dr/dφ) x (dr/dθ)| as a^2*sinφ. So, the surface area should be ∫∫a^2*sinφdφdθ.
I am trying to figure out the boundaries of integration. I see that the cylinder has the expression r = acosθ. So θ would be between –π/2 and π/2. But I am not sure what φ should be?

Alternately, I tried with cylindrical coordinates. So, I have that for the sphere I would use x = rcosθ, y = rsinθ, and z = a^2 - r^2. And if I let f be the vector function represented by x,y, and z, then df/dr = <cosθ, sinθ, -2r> and df/dθ = <-rsinθ, rcosθ, 0> And so if I take the cross product of the two vectors I would get 2r^2cosθ -2r^2sinθ +r. So |(df/dr)x(df/dθ)| is 4r^4 + r^2. So would the expression be ∫∫√(4r^4+r^2)drdθ where 0≤r≤rcosθ and –π/2≤θ≤π/2? Thanks!!
 

Answers and Replies

  • #2
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The problem with your assumption about the cylinder formula is that the one in your problem can have its axis somewhere other than a coordinate axis. What if you kept it in rectangular (x,y,z) coordinates?

The condition 0≤r≤rcosθ seems dubious....
 

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