# Thread: double integral question 2

1. ## double integral question 2

I = $\displaystyle \int$ between 0 to 1 dy, $\displaystyle \int$ 0 to y $\displaystyle y^3/(x^2 + y^2)$ dx

sketch the domain of integration

2. $\displaystyle \int_{0}^{1}{\int_{0}^{y}{\frac{y^{3}}{x^{2}+y^{2} }\,dx}\,dy}.$

The domain where the double integral is taken, it's a triangle.

3. Originally Posted by sonia1
I = $\displaystyle \int$ between 0 to 1 dy, $\displaystyle \int$ 0 to y $\displaystyle y^3/(x^2 + y^2)$ dx

sketch the domain of integration
The outer integral is form x= 0 to x= 1. Draw the vertical lines, x= 0, y= 1. The inner integral is from y= 0 to $\displaystyle y= x^3$. Draw that line and curve. The region of integration is the region bounded by those.

4. what do you when you evaluate it

5. You mean how do we evaluate this?

Reverse integration order first.