How do I evaluate a double integral with a trigonometric function?

math_04
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Homework Statement



Given the integral shown (in attachment), make a sketch of the region of integration, express the integral with the order of integration reversed and evaluate the integral after reversing the order of integration

Homework Equations





The Attempt at a Solution



So i got the limits as y<= x <= 1 and 0 <= y <= 1

But for the function y^2 sin xy dxdy, how should i integrate that. Integration by parts would not work because v= 0 so the whole integral becomes 0. Should i just read it off the table of integrals. if i do that, the integral becomes -y^2cosxy dx rite?

Thanks
 

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oh no, attachment pending approval. not again haha.
 
As said before, if you want quick replies, just host it on imageshack.us or some other image-hosting web service.
 
hahahaha~~~~~~~~

but why do they need to be approved?
 
You seem to have done the same thing you did with the previous problem: your "inner" integral goes from x to 1 while the shaded area in your picture is below the line y= x.
 
ohhh my god ur rite hallsofIvy thanks.

Alright, so the limits change but how do u evaluate that integral. Cant do it by parts cause of the reasons i mentioned.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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