Double Integral: Evaluate ∫∫(x^2 + y^2)dx dy in R

[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
then add them up.
Thanks

 

[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.

 

[mathsdespair]

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.

ok, it is y=-x+2

 

[mathsdespair]

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

 

[Mark44]

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

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?

 

[Ray Vickson]

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
then add them up.
Thanks

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.

 

[BiGyElLoWhAt]

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.

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.

 

[Ray Vickson]

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.

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

 

[mathsdespair]

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?

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?

 

[HallsofIvy]

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.

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

 

[mathsdespair]

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

is my method right for doing it into steps please?

 

[SammyS]

is my method right for doing it into steps please?

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?