Change of variables for double integral problem

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SUMMARY

The discussion focuses on the application of polar coordinates to evaluate the double integral of the function 1/sqrt(x^2+y^2) over a triangular region defined by the limits 0 < y < x and 0 < x < 3. The correct transformation involves using the equations x = rcos(θ) and y = rsin(θ), with the limits for θ set from 0 to π/4 and for r from 0 to 3sec(θ). The initial incorrect limits for r led to an erroneous result, which was clarified through community input.

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fuzzytoad
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Homework Statement


I want to use polar coordinates to integrate

1/sqrt(x^2+y^2) dydx with limits of integration 0 < y < x and 0 < x < 3

Homework Equations



x=rcosO
y=rsinO

The Attempt at a Solution



I know that the area being integrated over is the triangle enclosed by y=x and x=3. I have my limits for O to be 0 < O < pi/2, which I think is correct. I attempted to solve the problem using limits for r to be 3 < r < 3/cosO, however, the answer I obtain is 3(sqrt(2) - pi/2) whereas it should be 3ln(sqrt2 + 1). I think the problem is my limit for r.

Thanks in advance.

Edit: nevermind, I'm an idiot. Saw the problem right after I posted this.
 
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fuzzytoad said:

Homework Statement


I want to use polar coordinates to integrate

1/sqrt(x^2+y^2) dydx with limits of integration 0 < y < x and 0 < x < 3


Homework Equations



x=rcosO
y=rsinO


The Attempt at a Solution



I know that the area being integrated over is the triangle enclosed by y=x and x=3. I have my limits for O to be 0 < O < pi/2, which I think is correct. I attempted to solve the problem using limits for r to be 3 < r < 3/cosO, however, the answer I obtain is 3(sqrt(2) - pi/2) whereas it should be 3ln(sqrt2 + 1). I think the problem is my limit for r.

Thanks in advance.

Edit: nevermind, I'm an idiot. Saw the problem right after I posted this.
After converting to polar coordinates, your integral should be:
\int_{\theta = 0}^{\pi/4}\int_{r = 0}^{3sec(\theta)} \frac{r~dr~d\theta}{r}

Is that what you got?
 

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