In the future, please do not give answers to homework questions, even if you use a different approach from the one requested. See my earlier post in reply to this thread, and also review the forum FAQ, see below.
In the future, please do not give answers to homework questions, even if you use a different approach from the one requested. See my earlier post in reply to this thread, and also review the forum FAQ, see below.
Hum, is that formula for the surface area correct?
Technically it takes two integrals to find the surface area of a cylindar, but one is dr dt and the other is dh dt. If you do the dr and dh parts seperately then you end up with two dt integrals which can be combined since the limits of integration are the same for both. I thought you were having trouble with how to find the surface area of a cylindar using just calculus.
This is a Lagrange multiplier problem. As such it involves gradiants and solving a system of equations. There is no integration involved to solve this problem. Just derivatives. You are just finding where the gradiants point in the same direction and the constraint for the volume is satisfied.
If you plot V(r,h) and A(r,h) then you get a height field, i.e. z=f(x,y). At any given point on either surface the gradiant for that surface tells you what direction z increases most rapidly. Now if you plot a contour map of V you only care about points on one contour line. If you take any arbitrary point on that line then the gradiant for A is either orthogonal to it or isn''t. If it isn''t then there is some component of the gradiant tangent to the contour line. So you can move in that direction along the contour line and increase A. The gradiant for V at every point on that contour line is orthogonal to the contour line. If it wasn''t then you could move along the contour line and change the volume, but every point on the contour line is the same volume which is an obvious contridiction. That is why your gradiants have to point in the same opposite directions.
Now things are a little more complicated than that. You can''t really decide which way to go along the contour line based upon the angle between the tangent to the contour line and the gradiant of the function you are minimizing/maximizing. That would be the equivalent of using the slope of a y=f(x) function to decide what direction to go to the minimum. If you plot x*(x-1)*(x+1) you can see that clearly isn''t true. It does provide a way to visualize why your gradiants have to point the same or opposite directions at the maximum/minimum.
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In the future, please do not give answers to homework questions, even if you use a different approach from the one requested. See my earlier post in reply to this thread, and also review the forum FAQ, see below.
just in case anyone''s interested ive found the integral i wanted, integral of (RH + R^2) with respect to theta 0<theta<2pi. thanks for your suggestions though
