Double integral with absolute value

[norbellys]

Homework Statement

I am trying to evaluate double integral
∫∫D (|y - x²|)^½

D: -1<x<1, 0<y<2

Homework Equations

None

The Attempt at a Solution

I know that in order to integrate with the absolute value I have to split the integral into two parts:
y>x^2−−−>√y−x2
y>x^2−−−>√y−x2

I just can't get of the limits of the integral

(it is this same questions https://www.physicsforums.com/threads/absolute-value-in-a-double-integral.202157/ )

Please I been trying to set this up for hours how

 

[Mark44]

Homework Statement

I am trying to evaluate double integral
∫∫D (|y - x²|)^½

D: -1<x<1, 0<y<2

Homework Equations

None

The Attempt at a Solution

I know that in order to integrate with the absolute value I have to split the integral into two parts:
y>x^2−−−>√y−x2
y>x^2−−−>√y−x2

You've written the same thing twice. If y > x², then |y - x²| = y - x².
If y < x², then |y - x²| = -(y - x²) = x² - y.

I just can't get of the limits of the integral

Region D is just a rectangle.

(it is this same questions https://www.physicsforums.com/threads/absolute-value-in-a-double-integral.202157/ )

Please I been trying to set this up for hours how

 

[norbellys]

Homework Statement

I am trying to evaluate double integral
∫∫D (|y - x²|)^½

D: -1<x<1, 0<y<2

Homework Equations

None

The Attempt at a Solution

I know that in order to integrate with the absolute value I have to split the integral into two parts:
y>x^2−−−>√y−x2
y>x^2−−−>√y−x2

I just can't get of the limits of the integral

(it is this same questions https://www.physicsforums.com/threads/absolute-value-in-a-double-integral.202157/ )

Please I been trying to set this up for hours how

I forgot the function |y - x^2| is inside a square root. When I divide the two integrals for y > x2, |y - x2| = y - x2.
and y < x2, |y - x2| = -(y - x2) = x2 - y. I really don't know what would be the limits of integration would it be -1 to what then the other other what to 1?

 

[Charles Link]

I forgot the function |y - x^2| is inside a square root. When I divide the two integrals for y > x2, |y - x2| = y - x2.
and y < x2, |y - x2| = -(y - x2) = x2 - y. I really don't know what would be the limits of integration would it be -1 to what then the other other what to 1?

It takes a little practice to figure out how to select the limits of a double integral. If you choose to do the dy integral first, for each location x, the dy integral is performed and you need to figure out the limits (that often depend on x.). Then the dx integral is done adding up all the strips that were solved individually (as a function of x) when you did the dy integral.

 

[norbellys]

Thanks I think I got it!