# Double integral boundaries.

## Homework Statement

D = {x>0, x^2 < y < 10-x^2)

compute

integral (integral D of y^2 sqrt x)

## The Attempt at a Solution

I'm having trouble figuring out the bounds of the integral. y goes from x^2 to 10-x^2 but I think I have to split this integral up into two parts. I am not sure how to bound x. The parabolas intersect at the point sqrt 5

What do you mean probably at sqrt(5)? They intersect at two points

They intersect at (sqrt 5, 5) and (-sqrt 5, 5). x > 0 so we don't need the negative point.

Yes you got, it

So x goes from 0 to sqrt 5? thats the bound for x?

So x goes from 0 to sqrt 5? thats the bound for x?

What do you think it should be? Let's put it this way, if it isn't x = 0, where would you start from? You mentioned splitting the integral, how do you plan to do that?

I guess I'm overthinking. It looks like x goes from 0 to x^2 and then stops when x reaches sqrt 5, then goes from 10-x^2 back to 0. This is just from the drawing I mean.

Redbelly98
Staff Emeritus