I'm having trouble finding the answer to this problem. The square root is supposed to be over all of the x + y + 1.

Thanks.

Printable View

- Nov 10th 2008, 07:52 AMlarsonDouble Integration --- help!

I'm having trouble finding the answer to this problem. The square root is supposed to be over all of the x + y + 1.

Thanks. - Nov 10th 2008, 09:40 AMChris L T521

Note that when we integrate with respect to y, x+1 is considered a constant, let's say A.

The integral becomes

Now to evaluate this integral, you can make a simple substitution: . The limits of integration change as well! and

Thus, we now have

Now plugging back and simplifying, we end up with

I'm sure you can take it from here. :D

--Chris - Nov 10th 2008, 09:44 AMtbgh
I'm going to assume you're integrating by x and then by y. In this case it doesn't matter if it's the opposite, but sometimes it does.

The key is just remembering that the square root of a number is just an exponent of 1/2.

From here you're going to integrate as with any normal exponent. Add one to it and multiply the front by the inverse. Then just plug in 1 and 0 for x and subtract.

Just repeat for the other variable and you'll be there soon enough.