1. ## Multiple integral

How come I get a negative answer?

What's wrong?

2. Originally Posted by asi123
How come I get a negative answer?

What's wrong?

You do realise that $-1 \leq \sin r \leq 0$ for $\pi \leq r \leq 2 \pi$ .... ?

3. Originally Posted by mr fantastic

You do realise that $-1 \leq \sin r \leq 0$ for $\pi \leq r \leq 2 \pi$ .... ?
By $sin r$ you mean $\phi$?

like $0 \leq \phi \leq 2\pi$?

10x.

4. Originally Posted by asi123
By $sin r$ you mean $\phi$?

like $0 \leq \phi \leq 2\pi$?

10x.
The variable is r:

Your integrand is $r \sin r$ and you're integrating between $\pi \leq r \leq 2 \pi$. This integrand is always less than or equal to zero for $\pi \leq r \leq 2 \pi$. So it's unsurprising that the result of the integration is less than zero.

5. Got ya, 10x.

I have another one, how am I suppose to solve this one, I cant use polar coordinates, right?

I have no idea.

6. Originally Posted by asi123
Got ya, 10x.

I have another one, how am I suppose to solve this one, I cant use polar coordinates, right?

I have no idea.

You need to switch to a new set of coordinates. The integrand suggests the following transformation:

$u = x - y$ .... (1)

$v = x + y$ .... (2)

(1) and (2) imply $x = \frac{u + v}{2}$ and $y = \frac{v - u}{2}$.

Jacobian of the transformation: $J(u, v) = \frac{1}{2}$.

The region of integration in the xy-plane transforms into the region in the uv-plane bounded by the lines $v = u, ~ v = -u, ~ v = 1$ and $v = 2$.

So the integral becomes $\frac{1}{2} \int_{v=1}^{v=2} \int_{u=-v}^{u=v} e^{u/v} \, du \, dv$.