# Double integration

1. Apr 25, 2014

### mathsdespair

Evaluate ∫∫(x^2 + y^2)dx dy over the region enclosed within
R

(0,0), (2,0) and (1,1).
I am not asking someone to do the problem but to just verify, have I got the limits right?

I split it up into 2 legs
for the first leg integrate from , x: 0→1 and y :0→x
for the second leg integrate from x:1→2 and y:1→-x
Thanks

2. Apr 25, 2014

### BiGyElLoWhAt

First one looks good, I don't like the second one. Write the equation of the lin that goes from 1,1 to 2,0.

3. Apr 25, 2014

### mathsdespair

ok, it is y=-x+2

Last edited: Apr 25, 2014
4. Apr 25, 2014

### mathsdespair

so for the second leg is it x goes from 1 to 2 and y goes from 1 to -x+2?

5. Apr 25, 2014

### Staff: Mentor

No. The x values go from 1 to 2, as you said, but what do the y values do in the right-most triangle? Have you drawn a sketch of the region over which integration is taking place?

6. Apr 25, 2014

### Ray Vickson

You are doing the y-integration first; that is, for each fixed x you first integrate over y. The alternative is to do the x-integration first; that is, for each fixed y, first integrate over x. You might find this easier.

7. Apr 25, 2014

### BiGyElLoWhAt

Why?
A)you solve for y in terms of x,
B)you solve for x in terms of y,

Seems pretty tomato tomatoe to me.

8. Apr 25, 2014

### Ray Vickson

The best way to know why is to try it for yourself.

9. May 17, 2014

### mathsdespair

so for the first leg
x:0 to 1
y: 0 to x

2nd leg
x:1 to 2
y:1 to -x+2
is that right?

10. May 17, 2014

### HallsofIvy

Staff Emeritus
Because you could do it in a single integration rather than needing to break it up into two integrals.

11. May 17, 2014

### mathsdespair

is my method right for doing it in to steps please?

12. May 17, 2014

### SammyS

Staff Emeritus
No. The method may be right, but your implementation of the method for doing it in steps is wrong.

What is the upper boundary of the right hand region? (the one from x=1 to x=2)

What is the lower boundary of the right hand region?