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mulitvariablecalcuser
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Hi, I am trying to find the double Integral of [(1 + x + y)^.5, {x, 0, 1}, {y, 0, 2}] do you switch to cylindrical coordinates?
 one year ago
 one year ago
mulitvariablecalcuser Group Title
Hi, I am trying to find the double Integral of [(1 + x + y)^.5, {x, 0, 1}, {y, 0, 2}] do you switch to cylindrical coordinates?
 one year ago
 one year ago

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Waynex Group TitleBest ResponseYou've already chosen the best response.1
I'm not an expert on this. I don't recognize that form of equation. Could you elaborate a bit on what the parameterization is intended to do there? Or perhaps which portion of the class this came from.
 one year ago

mulitvariablecalcuser Group TitleBest ResponseYou've already chosen the best response.0
It is from the chapter on The change of variables theorem. The full question states Evaluate the double integral dxdy/sqrt(1+x+2y) on the region D, where D=[0,1]*[0,1], by setting T(u,v)=(u,v/2) and evaluating an integral D*, where T(D*)=D. I found that the jacobian is 1/2 and the new integral is 1/sqrt(1+u+v) dudv, where u is between 0 and 1 and v is between 0 and 2. I just don't remember how to integrate the negative square root of (1+u+v).
 one year ago

mulitvariablecalcuser Group TitleBest ResponseYou've already chosen the best response.0
Both my solutions manual and mathematica say the answer is 2/3[(92*(sqrt2)3*(sqrt3))] and that D is the region 0≤u≤1 and 0≤v≤2
 one year ago

Waynex Group TitleBest ResponseYou've already chosen the best response.1
Thanks for the extra information. Normally, when changing variables, if one were to use a change of variables involving cylindrical coordinates, we would change immediately from x and y to r and theta instead of x and y to u and v, and then to r and theta. In this case, integrating 1 over the square root of 1 + u + v is rather straightforward. Consider 1 over the square root of u. It's almost the same as 1 over the square root of 1 + u. I'll use h instead of u, to avoid confusion. Substitute h = 1+u, and dh = du. Hm, well then we are back to the simple case of 1/sqrt(h). A better substitution then is h = 1 + v + u and dh = du. And again we are back to 1/sqrt(h). I hope that helps.
 one year ago

mulitvariablecalcuser Group TitleBest ResponseYou've already chosen the best response.0
This was totally helpful. Thank you!
 one year ago

Waynex Group TitleBest ResponseYou've already chosen the best response.1
You're welcome ;)
 one year ago
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