A community for students.
Here's the question you clicked on:
 0 viewing
mulitvariablecalcuser
 one year 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?
mulitvariablecalcuser
 one year 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?

This Question is Closed

Waynex
 one year 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.

mulitvariablecalcuser
 one year 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).

mulitvariablecalcuser
 one year 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
 one year 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.

mulitvariablecalcuser
 one year ago
Best ResponseYou've already chosen the best response.0This was totally helpful. Thank you!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.