Trying to find this double integral using polar coordinates

[devinaxxx]

Homework Statement

question :
find the value of
$\iint_D \frac{x}{(x^2 + y^2)}dxdy$

domain : 0≤x≤1,x2≤y≤x

Homework Equations

The Attempt at a Solution

so here, i tried to draw it first and i got that the domain is region in first quadrant bounded by y=x2 and y=x

and i decided to convert the equation to polar coordinate,

$\int_{0}^{\pi/2}\int_{(r^2cos^2\theta)}^{rcos \theta}cos \theta dr d\theta\\$
i proceed to solve the integral, but still got r in the last equation. the answer of the book said 1/2log2
and also i already tried without using polar coordinate and just using the cartessian and i got 1/2log2, but I am so curious can i do this problem using polar coordinate or not? thanks

 

[Mark44]

Homework Statement

question :
find the value of
$\iint_D \frac{x}{(x^2 + y^2)}dxdy$

domain : 0≤x≤1,x2≤y≤x

I believe you mean the 2nd inequality to say $x^2 \le y \le x$. If so, at the very least, write x^2 as you did above.

Homework Equations

The Attempt at a Solution

so here, i tried to draw it first and i got that the domain is region in first quadrant bounded by y=x2 and y=x

$y = x^2$?

and i decided to convert the equation to polar coordinate,

$\int_{0}^{\pi/2}\int_{(r^2cos^2\theta)}^{rcos \theta}cos \theta dr d\theta\\$

Your polar integral doesn't look right to me. The outer integral should be from $\theta = 0$ to $\theta = \pi/4$. For the inner integral, r ranges from 0 to the graph of $y = x^2$. When this equation is converted to polar form, the first result would be $r\cos(\theta) = r^2 \cos^2(\theta)$. Also, dxdy gets converted to $r dr d\theta$, but I can't tell from your work whether you did that.

i proceed to solve the integral, but still got r in the last equation. the answer of the book said 1/2log2

What you wrote would normally be interpreted as $\frac 1 2 \log(2)$. Do you mean $\frac 1 {2 \log(2)}$? Also, many textbooks use log(x) to mean the common log, or log base-10, and use ln(x) for the natural log.

and also i already tried without using polar coordinate and just using the cartessian and i got 1/2log2, but I am so curious can i do this problem using polar coordinate or not? thanks

 

[Dick]

but I am so curious can i do this problem using polar coordinate or not? thanks

Yes, you can do it using polar coordinates, but your limits are all wrong. Draw a sketch of the region and rethink them.

 

[devinaxxx]

I believe you mean the 2nd inequality to say $x^2 \le y \le x$. If so, at the very least, write x^2 as you did above.
$y = x^2$?

Your polar integral doesn't look right to me. The outer integral should be from $\theta = 0$ to $\theta = \pi/4$. For the inner integral, r ranges from 0 to the graph of $y = x^2$. When this equation is converted to polar form, the first result would be $r\cos(\theta) = r^2 \cos^2(\theta)$. Also, dxdy gets converted to $r dr d\theta$, but I can't tell from your work whether you did that.
What you wrote would normally be interpreted as $\frac 1 2 \log(2)$. Do you mean $\frac 1 {2 \log(2)}$? Also, many textbooks use log(x) to mean the common log, or log base-10, and use ln(x) for the natural log.

hello thankyou for answering my question, and i will edit my post. i think i got it that the outer coordinate is, $\theta = 0$ to $\theta = \pi/4$ by using coordinate $(1,1)$ and $tan \theta = y/x$ right? and for the r itself, is it $\sqrt 2 \le r \le \sqrt 2$? and also i was confused, i drawed in cartessian coordinate,after i changed to polar coordinate, is it always going to be a circle?

 

[devinaxxx]

Homework Statement

question :
find the value of
$\iint_D \frac{x}{(x^2 + y^2)}dxdy$

domain :$0≤x≤1,x^2≤y≤x$

Homework Equations

The Attempt at a Solution

so here, i tried to draw it first and i got that the domain is region in first quadrant bounded by $y=x^2$ and $y=x$

and i decided to convert the equation to polar coordinate,

$\int_{0}^{\pi/2}\int_{(r^2cos^2\theta)}^{rcos \theta}cos \theta dr d\theta\\$
i proceed to solve the integral, but still got r in the last equation. the answer of the book said $\frac{1}{2}log2$
and also i already tried without using polar coordinate and just using the cartessian and i got 1/2log2, but I am so curious can i do this problem using polar coordinate or not? thanks

 

[devinaxxx]

Yes, you can do it using polar coordinates, but your limits are all wrong. Draw a sketch of the region and rethink them.

hello thankyou for answering my question, i got it that the outer coordinate is, $\theta = 0$ to $\theta = \pi/4$ by using coordinate $(1,1)$ and $tan \theta = y/x$ right? and for the r itself, is it $\sqrt 2 \le r \le \sqrt 2$ or $rcos \theta \le r \le 1$,, is this true??

 

[Dick]

hello thankyou for answering my question, i got it that the outer coordinate is, $\theta = 0$ to $\theta = \pi/4$ by using coordinate $(1,1)$ and $tan \theta = y/x$ right? and for the r itself, is it $\sqrt 2 \le r \le \sqrt 2$ or $rcos \theta \le r \le 1$,, is this true??

The $\theta$ range is right. The outer value of $r$ will depend on $\theta$, won't it?? It's $\sqrt 2$ at $\theta = \pi/4$ but decreases to zero at $\theta=0$. When you write $r$ as a function of $\theta$ check those two conditions.

 

[devinaxxx]

The $\theta$ range is right. The outer value of $r$ will depend on $\theta$, won't it?? It's $\sqrt 2$ at $\theta = \pi/4$ but decreases to zero at $\theta=0$. When you write $r$ as a function of $\theta$ check those two conditions.

okay i think, $0 \le r \le tan \theta sec \theta$ ? since the region is from origin to $y= x^2$ ?

 

[Dick]

okay i think, $0 \le r \le tan \theta sec \theta$ ? since the region is from origin to $y= x^2$ ?

You've got it.

 

[devinaxxx]

Thankyou so much but in this question can i just substitue directly to the x range? So $0 \le x \le 1 =0 \le r sin /theta \le 1$?

 

[devinaxxx]

You've got it.

Thankyou so much but can i just subtitue directly to $0 \le x \le 1 = 0 \le rsin \theta \le 1$ to get the r value? Or i should draw it?

 

[Dick]

Thankyou so much but can i just subtitue directly to $0 \le x \le 1 = 0 \le rsin \theta \le 1$ to get the r value? Or i should draw it?

I'm not sure why you are asking. Didn't we decide that the ranges are $0 \le \theta \le \pi/4$ and $0 \le r \le \sec \theta \tan \theta$? Isn't that all we need to know?

 

[devinaxxx]

I'm not sure why you are asking. Didn't we decide that the ranges are $0 \le \theta \le \pi/4$ and $0 \le r \le \sec \theta \tan \theta$? Isn't that all we need to know?

yes but I am just wondering can i just substitute to the domain in cartessian to find the range?

 

[Mark44]

yes but I am just wondering can i just substitute to the domain in cartessian to find the range?

I think you are confused about the meaning of these terms. In the context of this problem, domain means the region in the plane over which integration is being performed. The domain is described in terms of intervals (probably a better choice of words than "range").

In Cartesian form, the region of integration is described by the pair of inequalities $0 \le x \le 1; x^2 \le y \le x$, and the integrand is in Cartesian form.

When you convert an integral to polar form, you have to convert everything to polar form: the integrand, the description of the region over which integration is being performed (which determines what the limits of integration will be), and the differentials (i.e., dxdy or dydx becomes $rdrd\theta$.

You posted this problem 11 days ago. What's keeping you from finishing it?

 

[LCKurtz]

yes but I am just wondering can i just substitute to the domain in cartessian to find the range?

You know what the correct limits for the polar integral are. Are you asking if you can get the polar coordinate limits by substituting directly into the xy double integral limits? Have you tried doing that? Did it work? You should be able to tell us whether or not you can do it.