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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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Waynex
 3 years ago
Best ResponseYou've already chosen the best response.1I'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It 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).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Both 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

Waynex
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This was totally helpful. Thank you!
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