Multiple integral

**asi123** · July 16th 2008, 02:30 AM

How come I get a negative answer?

What's wrong?

10x in advance.

**mr fantastic** · July 16th 2008, 02:39 AM

Originally Posted by asi123

How come I get a negative answer?

What's wrong?

10x in advance.

Your answer is correct.

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

**asi123** · July 16th 2008, 03:09 AM

Originally Posted by mr fantastic

Your answer is correct.

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.

**mr fantastic** · July 16th 2008, 03:16 AM

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.

**asi123** · July 16th 2008, 03:44 AM

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.

10x in advance.

**mr fantastic** · July 16th 2008, 04:10 AM

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.

10x in advance.

New questions require new threads.

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

Thread: Multiple integral

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