Does it look like I'm doing this double integral correctly?

[DottZakapa]

Homework Statement

{(x,y)∈R^2|x^2+y^2<=4, y<=x+2 }

integral of 8x dx dy

Relevant Equations

double integral

are the boundaries of integration correct?
i split the domain in two as follows

-2<=x<=0 , -(4-x^2)^(1/2)<y<=x+2 and
0<=x<=2 -(4-x^2)^(1/2)<=y<=(4-x^2)^(1/2)

 

[PeroK]

Looks all right.

 

[DottZakapa]

if, concerning the one on the right, if I use polar coordinates, the boundaries of integration become
0<=r<= 2 and pi/2 <= $\theta$<=-pi/2

is it correct doing so?

 

[PeroK]

if, concerning the one on the right, if I use polar coordinates, the boundaries of integration become
0<=r<= 2 and pi/2 <= $\theta$<=-pi/2

is it correct doing so?

Yes, although you could include the third quadrant as well in that case: $-\pi \le \theta \le \frac \pi 2$. And that simplifies the bounds on the other integral.

 

[DottZakapa]

so i don't understand why my result doesn't match with the solution :(

 

[PeroK]

so i don't understand why my result doesn't match with the solution :(

What is your answer?

 

[DottZakapa]

-32 the one on the left and -128/3 the one with polar coordinates

 

[PeroK]

-32 the one on the left and -128/3 the one with polar coordinates

The answer must be positive, as there is more positive $x$ than negative $x$ remaining. I think you got a sign wrong somewhere.

 

[DottZakapa]

mind if i post the picture of my calculations? just to save typing time?

The answer must be positive, as there is more positive $x$ than negative $x$ remaining. I think you got a sign wrong somewhere.

Do you mind if a post the picture of my calculations? just to save time

 

[PeroK]

mind if i post the picture of my calculations? just to save typing time?

Do you mind if a post the picture of my calculations? just to save time

Okay by me.

 

[PeroK]

Just an idea, did you integrate $\cos \theta$ as $-\sin \theta$?

 

[DottZakapa]

 

[PeroK]

You've got the bounds on the integral the wrong way round.

 

[DottZakapa]

could be that being integrating from pi/2 to -pi/2 should i put a minus sign in front of the integral due to the clock wise direction ?

 

[DottZakapa]

You've got the bounds on the integral he wrong was round.

sorry I'm not getting what you are saying

 

[PeroK]

could be that being integrating from pi/2 to -pi/2 should i put a minus sign in front of the integral due to the clock wise direction ?

The way $\theta$ is defined in polar coordinates, you integrate anti-clockwise. In this case from $-\frac \pi 2$ to $\frac \pi 2$.

 

[PeroK]

could be that being integrating from pi/2 to -pi/2 should i put a minus sign in front of the integral due to the clock wise direction ?

Yeah, but you didn't integrate $r$ from $2$ to $0$ and put a minus sign in front!

 

[DottZakapa]

ok integrating from -pi/2 to pi/2 and putting al together everything works thanks

 

[Mark44]

You've got the bounds on the integral he wrong was round.

A couple typos here might have confused the OP. This should read "You've got the bounds on the integral the wrong way round."

@DottZakapa, keep in mind that $\int_a^b f(x) dx = -int_b^a f(x) dx$, so if you integrate in the reverse direction, the sign of the result changes. That's what @PeroK was talking about in the quote above.