Recent content by mman014

Disregard what I had here before, misread what time you were trying to find. Not sure what other way to do it.

Anyways thanks for making me realize my mistake haha. I was staring at the problem for too long, fresh eyes always help.

Wow so I just realized the mistake I made, wasn't thinking properly., equation of the line is just y = 2x so x = .5y, and I got it from there just using u sub and the rest is easy.

Thanks. Anyways so I figured out that I had the limits wrong, and when I change it I am getting ∫^{1}_{0}∫^{sin(∏*y^2)}_{0}sin(∏*y^2)dxdy So from there I get ∫^{1}_{0}sin(∏*y^2)[x]^{sin(∏*y^2)}_{0} = ∫^{1}_{0} sin^{2}(∏*y^2) dy At this point I am stuck again, unless I am doing something wrong...

Homework Statement Let f(x,y) = sin(∏*y^2). Let R be the triangular region on the x-y plane with the vertices at (0,0) (0,1) (.5,1). Consider the solid that is under z = f(x,y) and over the region R. Calculate the volume over that region using double integrals. Homework Equations The...