# Evaluating Cartesian integral in polar coordinates

• toforfiltum
In summary: No. Answer the two questions I asked before (now in red). You have about 10 minutes before I hit the sack. [Edit:] I see you corrected it so now it's correct. Write it as ##3\sec\theta## and integrate...Ah, I see. So the bounds are actually ##0 \leq r \leq \frac{3}{cos\theta}##.Thank you so much for all the help!Ah, I see. So the bounds are actually ##0 \leq r \leq \frac{3}{cos\theta}##.Thank you so much for all the help!
toforfiltum

## Homework Statement

Transform given integral in Cartesian coordinates to one in polar coordinates and evaluate polar integral.

##\int_{0}^3 \int_{0}^x \frac {dydx}{\sqrt(x^2+y^2)}##

## The Attempt at a Solution

I drew out the region in the ##xy## plane and I know that ##0 \leq \theta \leq \frac{\pi}{4}##.
For ##r##, I thought that it should be ##3 \leq r \leq 3\sqrt 2##

So my polar integral is ##\int_{0}^ \frac{\pi}{4} \int_{3}^{3\sqrt 2} \frac{1}{\sqrt(r^2)} r drd\theta##.

The answer I get from this integral is ##\frac{3\pi}{4} (\sqrt2 -1)##

But the answer is ##3 \ln(\sqrt2+1)##.

I have no idea how ##ln## appears in the answer.

Where am I wrong?

Thanks!

toforfiltum said:

## Homework Statement

Transform given integral in Cartesian coordinates to one in polar coordinates and evaluate polar integral.

##\int_{0}^3 \int_{0}^x \frac {dydx}{\sqrt(x^2+y^2)}##

## The Attempt at a Solution

I drew out the region in the ##xy## plane and I know that ##0 \leq \theta \leq \frac{\pi}{4}##.
For ##r##, I thought that it should be ##3 \leq r \leq 3\sqrt 2##

So my polar integral is ##\int_{0}^ \frac{\pi}{4} \int_{3}^{3\sqrt 2} \frac{1}{\sqrt(r^2)} r drd\theta##.

The answer I get from this integral is ##\frac{3\pi}{4} (\sqrt2 -1)##

But the answer is ##3 \ln(\sqrt2+1)##.

I have no idea how ##ln## appears in the answer.

Where am I wrong?

Thanks!
Assuming you are working the same problem as appears in your book, I agree with you. If this is the problem --
##\int_{0}^3 \int_{0}^x \frac {dydx}{x^2+y^2}##
that would explain the book's answer.

Mark44 said:
Assuming you are working the same problem as appears in your book, I agree with you. If this is the problem --
##\int_{0}^3 \int_{0}^x \frac {dydx}{x^2+y^2}##
that would explain the book's answer.
Ah, thank you! I did check the question again in the book, and I did not misread it.

Mark44 said:
Assuming you are working the same problem as appears in your book, I agree with you. If this is the problem --
##\int_{0}^3 \int_{0}^x \frac {dydx}{x^2+y^2}##
that would explain the book's answer.
toforfiltum said:

## Homework Statement

Transform given integral in Cartesian coordinates to one in polar coordinates and evaluate polar integral.

##\int_{0}^3 \int_{0}^x \frac {dydx}{\sqrt(x^2+y^2)}##

## The Attempt at a Solution

I drew out the region in the ##xy## plane and I know that ##0 \leq \theta \leq \frac{\pi}{4}##.
For ##r##, I thought that it should be ##3 \leq r \leq 3\sqrt 2##

LCKurtz said:
Your limits for ##r## are incorrect.
Are both upper and lower ones incorrect? I don't know any other way to find the limits for ##r##. Any hints?

Or since value of ##x## is constant, are the bounds in Cartesian coordinates something like ##3 \leq \sqrt{9+y^2} \leq 3\sqrt2##?

toforfiltum said:
Are both upper and lower ones incorrect? I don't know any other way to find the limits for ##r##. Any hints?

Or since value of ##x## is constant, are the bounds in Cartesian coordinates something like ##3 \leq \sqrt{9+y^2} \leq 3\sqrt2##?
Both are incorrect. What does the region look like? Put your pencil on the origin and move it in the ##r## direction. What does it hit?

LCKurtz said:
Both are incorrect. What does the region look like? Put your pencil on the origin and move it in the ##r## direction. What does it hit?
I honestly do not know. In the ##xy## plane, the region is bounded by the line ##y=x##, the ##x## axis and the line ##x=3##. What does moving it in the ##r## direction mean? If I do so, for ##\theta=0##, I get ##0 \leq r \leq 3##, but for ##\theta=\frac{\pi}{4}##, I get ##0 \leq r \leq 3 \sqrt2##.

I'm confused.

toforfiltum said:
I honestly do not know. In the ##xy## plane, the region is bounded by the line ##y=x##, the ##x## axis and the line ##x=3##. What does moving it in the ##r## direction mean? If I do so, for ##\theta=0##, I get ##0 \leq r \leq 3##, but for ##\theta=\frac{\pi}{4}##, I get ##0 \leq r \leq 3 \sqrt2##.

I'm confused.
Well, at least you are thinking about it. You have noticed that if ##\theta = 0##, ##r## goes from ##0## to ##3## and for ##\theta = \frac \pi 4##, ##r## goes from ##0## to ##3\sqrt 2##. Both observations are correct. So what you have noticed is what ##r## goes to depends on what ##\theta## is. So if you take some in between value of ##\theta## and move ##r## in that direction, don't you hit the line ##x=3##? What is the equation of that in polar coordinates? What is ##r## on that line? It will depend on ##\theta##.

LCKurtz said:
Well, at least you are thinking about it. You have noticed that if ##\theta = 0##, ##r## goes from ##0## to ##3## and for ##\theta = \frac \pi 4##, ##r## goes from ##0## to ##3\sqrt 2##. Both observations are correct. So what you have noticed is what ##r## goes to depends on what ##\theta## is. So if you take some in between value of ##\theta## and move ##r## in that direction, don't you hit the line ##x=3##? What is the equation of that in polar coordinates? What is ##r## on that line? It will depend on ##\theta##.
Ah, I see. Is it ##0 \leq r \leq \frac{3}{cos\theta}##?

LCKurtz said:
Well, at least you are thinking about it. You have noticed that if ##\theta = 0##, ##r## goes from ##0## to ##3## and for ##\theta = \frac \pi 4##, ##r## goes from ##0## to ##3\sqrt 2##. Both observations are correct. So what you have noticed is what ##r## goes to depends on what ##\theta## is. So if you take some in between value of ##\theta## and move ##r## in that direction, don't you hit the line ##x=3##? What is the equation of that in polar coordinates? What is ##r## on that line? It will depend on ##\theta##.

toforfiltum said:
Ah, I see. Is it ##0 \leq r \leq \frac{3}{cos\theta}##?

No. Answer the two questions I asked before (now in red). You have about 10 minutes before I hit the sack. [Edit:] I see you corrected it so now it's correct. Write it as ##3\sec\theta## and integrate it.

LCKurtz said:
No. Answer the two questions I asked before (now in red). You have about 10 minutes before I hit the sack. [Edit:] I see you corrected it so not it's correct. Write it as ##3\sec\theta## and integrate it.
Thanks! if it weren't correct, it would be too much pressure for me. 10 minutes!

Anyway, thanks so much for your time, and good night!

Apologies for my incorrect conclusion. r indeed does depend on ##\theta##.

## 1. What is the difference between Cartesian and polar coordinates?

Cartesian coordinates are a system of representing points in a plane using the x and y coordinates, while polar coordinates use the distance from the origin and the angle from the positive x-axis to represent points.

## 2. Why would you need to evaluate a Cartesian integral in polar coordinates?

Some integrals can be more easily evaluated in polar coordinates, especially if the region being integrated over has a circular or radial symmetry. Additionally, some physical problems are best described using polar coordinates.

## 3. How do you convert a Cartesian integral to a polar integral?

To convert a Cartesian integral to a polar integral, you need to substitute the x and y coordinates in the integral with their equivalent expressions in terms of r and θ. This can be done using the following equations: x = rcosθ and y = rsinθ.

## 4. What are the limits of integration in a polar integral?

In a polar integral, the limits of integration for r typically range from 0 to some value representing the radius of the region being integrated over. The limits for θ can vary depending on the shape of the region, but they are typically 0 to 2π for a full circle, or 0 to π for a half-circle.

## 5. Are there any advantages to using polar coordinates over Cartesian coordinates?

Yes, there are several advantages to using polar coordinates over Cartesian coordinates. Some integrals can be more easily evaluated in polar coordinates, and certain physical problems are best described using polar coordinates. Additionally, polar coordinates can simplify certain geometric constructions and make it easier to visualize and understand certain concepts in mathematics.

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