At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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.
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).
Not the answer you are looking for? Search for more explanations.
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.