Recent content by vandyboy73191

by the way. thank you so much. your a life saver

holy sh*t. lol wow that was eye opening. so above y=x within the circle y is greater than x. Below y=x x is greater than y. This means that the integral of the top half is the opposite of the integral of the bottom half and thus they sum to 0. Wow. That was cool.

the statement is correct. my professor says its a trick question that can not be solved by straight integrating. Just to reiterate it is over the domain of a disk of radius 2 and center (1,1). I reached the conclusion that its not integrable about 1.5 hrs ago, but I can't find this trick he's...

http://i.imagehost.org/view/0153/mathproblem3 that is the original integrand. I basically just substituted x=rcos(theta) and y=rsin(theta) and multiplied by r to convert to polar. But I don't know what to do after that

Homework Statement Evaluate the double integral sin(x-y)*e(x-y)^2-0y) 2--- dA where D is a disk of radius 2 whose center is (1; 1) Homework Equations The Attempt at a Solution gee this...

Oops. Yeah I just noticed that. thanks guys

Homework Statement 1. Find the volume of the solid which is under the surface z = 2x + y2 and above the region bounded by x = y^2 and x = y^3. Homework Equations The Attempt at a Solution So first I graphed x=y^3 and x=y^2. (http://h.imagehost.org/view/0716/Math_Problem ) I found their...