Question

Double integral problem, posted below shortly.

Answer 1

\[\text{Evaluate the double integral} \iint \frac{y^3}{x^2+y^2}dA \ \text{where A is the triangle with}\]\[\text{vertices (0,0), (0,1), and (1,1).}\]

Answer 2

@ganeshie8

Answer 3

Alright, so these ones should be more straightforward and hopefully my saving grace. That is, assuming he doesn't give us any awful integration. Bounds are established by the object A.

Answer 4

\[\iint \frac{y^3}{x^2+y^2}dA=\int\limits_{0}^{1}\int\limits_{0}^{x}\frac{y^3}{x^2+y^2}dydx\]

Answer 5

I'm wondering whether I should try to switch the order of integration or something, this whole thing looks kind of unpleasant, but I can maybe see the advantages of integrating w.r.t. x first?

Answer 6

\[\int\limits_{0}^{1}\int\limits_{y}^{1}\frac{y^3}{x^2+y^2}dxdy\]

Answer 7

No, nevermind, I'm not sure how to proceed with this integration-wise, I should know how, but I don't, save a substitution, which is god-awful. How should I approach this integral?

Answer 8

(?)

Answer 9

Since I'm treating y like a constant, could I integrate this using a trig formula where the denominator is like 1 + x^2, and the numerator's y^3 is factored out of the integral?

Answer 10

@Kainui

Answer 11

@perl

Answer 12

@zepdrix

Answer 13

@hartnn

Answer 14

are we talking about this :
\(\int\limits_{0}^{1}\int\limits_{y}^{1}\frac{y^3}{x^2+y^2}dxdy\)

yes, treat y as constant and bring y^3 outside inner integral

Answer 15

inner integral will be of the form 1/(x^2+a^2)

Answer 16

Alright, cool. I think I'm not going to proceed forward with that, because I don't think he'll give something like that on our final, lol...

Answer 17

But, my setup was correct, right? Bounds of integration-wise?

Answer 18

checking

Answer 19

yep, bounds are correct :)

Answer 20

I believe the bounds of integration are off: