Calc 3 - Double Integral (-2y lnx) dA, R bounded by y = 4...

[Jamers328]

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region R>

Double Integral (-2y lnx) dA
R: region bounded by y=4-x^2, y=4-x

I set it up as the integral from 0 to 1, and the integral from 4-x to 4-x^2 of (-2y lnx) dy dx
The answer is 26/25. I am not doing something right with the work though!

 

[galactus]

You have a convoluted post. Typos.

I think you mean: \(\displaystyle \int_{0}^{1}\int_{4-x}^{4-x^{2}}[-2yln(x)]dydx=\frac{26}{25}\)

But you posted y=4 and y=4-x. If you use the ones you posted, you will not get 26/25. Is this correct?.

Reverse integration order:

\(\displaystyle \int_{3}^{4}\int_{4-y}^{\sqrt{4-y}}[-2yln(x)]dxdy=\frac{26}{25}\)

 

[Jamers328]

I'm sorry. I changed it, you were right.