# Find the double integral

• Mar 2nd 2010, 07:58 AM
Rhode963
Find the double integral
So I'm confused on how to take the first integral because of the radical... please explain how you came to your answer. (S=integral, so S:0,1 means the integral from 0 to 1, just to explain my notation)

S:0,1(S:0,1 $(xy((x^2+y^2)^(1/2)))$dy)dx
• Mar 2nd 2010, 08:15 AM
TheEmptySet
Quote:

Originally Posted by Rhode963
So I'm confused on how to take the first integral because of the radical... please explain how you came to your answer. (S=integral, so S:0,1 means the integral from 0 to 1, just to explain my notation)

S:0,1(S:0,1 $(xy((x^2+y^2)^(1/2)))$dy)dx

I think this is what you mean

$\int_{0}^{1}\int_{0}^{1}\frac{xy}{\sqrt{x^2+y^2}}d ydx$

It is just a u sub. let $u=x^2+y^2 \implies du=2ydy$ Remember x is constant with respect to the inside integral.

This is exactly like integrating

$\int_{0}^{1}\frac{3y}{\sqrt{3^2+y^2}}dy$
where three is playing the role of x!

I hope this helps.
• Mar 2nd 2010, 08:18 AM
Rhode963
Great help, but it is actually the double integral of $xy*sqrt(x^2+y^2)$ from 0 to 1. Sorry, I don't know how to get the integral symbol to work.
• Mar 2nd 2010, 08:29 AM
Rhode963
Would u-sub still work? So for the inner integral I could pull out an x, and get x times the integral of $sqrt(u)du$?
• Mar 2nd 2010, 08:31 AM
TheEmptySet
Quote:

Originally Posted by Rhode963
Would u-sub still work? So for the inner integral I could pull out an x, and get x times the integral of $sqrt(u)du$?

It doesn't change anything the substituion still works.
• Mar 2nd 2010, 08:35 AM
Rhode963
So for the inside integral, I got the answer to be $(3/2)x$, does that sound right to you? It seems like a strange amount of canceling. And when plugging into the final integral, I got the final answer to be 3.
• Mar 2nd 2010, 01:09 PM
General
Quote:

Originally Posted by Rhode963
Great help, but it is actually the double integral of $xy*sqrt(x^2+y^2)$ from 0 to 1. Sorry, I don't know how to get the integral symbol to work.

$\int_0^1 \int_0^1 xy \, \sqrt{x^2+y^2} dy \, dx =\int_0^1 x \int_0^1 y \, \sqrt{x^2+y^2} dy \, dx$.

Just apply the suggested substitution on the inner integral.