Evaluating Cartesian integral in polar coordinates

[toforfiltum]

Homework Statement

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

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

Homework Equations

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!

 

[Mark44]

Homework Statement

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

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

Homework Equations

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.

 

[toforfiltum]

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.

 

[LCKurtz]

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.

Homework Statement

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

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

Homework Equations

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 $3 \leq r \leq 3\sqrt 2$

Your limits for $r$ are incorrect. Your text's answer is correct for the problem as stated.

 

[toforfiltum]

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 $3 \leq \sqrt{9+y^2} \leq 3\sqrt2$?

 

[LCKurtz]

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?

 

[toforfiltum]

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.

 

[LCKurtz]

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 $\theta$.

 

[toforfiltum]

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]

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 $0 \leq r \leq \frac{3}{cos\theta}$?

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.

 

[toforfiltum]

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!

 

[Mark44]

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