Region bounded by a line and a parabola (polar coordinates)

[Poetria]

Homework Statement

$r=\frac 1 {cos(\theta)+1}$
y=-x

A region bounded by this curve and parabola is to be found.

2. The attempt at a solution

I have found the points of intersection but I am not sure what to do with the line (I need polar coordinates and it is not dependent on r :( )

This is my solution (but it is wrong :( )

integral [-pi/4, 3*pi/4] (1/2*((1/(cos(theta)+1))^2-(-pi/4)^2))

 

[BvU]

Did you make a sketch ?
A line like y = -x is simply described as $\theta = -{\pi\over 4}$ in polar coordinates ...

Now that you have found the region, are you done ? Or was the problem statement that you had to find the area as well ?

 

[Poetria]

Did you make a sketch ?
A line like y = -x is simply described as $\theta = -{\pi\over 4}$ in polar coordinates ...

Now that you have found the region, are you done ? Or was the problem statement that you had to find the area as well ?

The task is to find an integral with boundaries (area of the region) - it is not required to compute it.
Well, I understand that $\theta= \frac {-\pi} 4$ hm but something is wrong with my solution.
Oh yes, I made a sketch. :(

Another piece of information: limits of integration in the range (-pi, pi).

 

[BvU]

I made a sketch. :(

Can't see it. Mine looks just fine. What are you going to integrate and with what bounds ?

 

[BvU]

Another piece of information: limits of integration in the range (-pi, pi).

Big deal. We already have this $-{\pi\over 4}$ and the other point can only have $+{3\pi\over 4}$ because it is on the line too...

 

[Poetria]

Can't see it. Mine looks just fine. What are you going to integrate and with what bounds ?

This is the integral I already posted (but not in Latex):

integral [-pi/4, 3*pi/4] (1/2*((1/(cos(theta)+1))^2-(-pi/4)^2))
bounds: $-\frac \pi 4, \frac {3*\pi} 4$

 

[BvU]

Ok, so the bounds are fine.

(1/2*((1/(cos(theta)+1))^2-(-pi/4)^2))

What's that ?

 

[Poetria]

Big deal. We already have this $-{\pi\over 4}$ and the other point can only have $+{3\pi\over 4}$ because it is on the line too...

Ha ha right but I am maniacally checking everything.

 

[BvU]

Study https://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_(area)

(either that or transform to an integral in y and x)

 

[Poetria]

Ok, so the bounds are fine.
What's that ?

dA=1/2*r^2 dtheta

So I tried to subtract a piece of the parabola squared from the line (but here I have a problem - no r).

 

[Poetria]

Study https://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_(area)

(either that or transform to an integral in y and x)

Oh OK, I will take a look at it. :)

 

[Poetria]

Study https://en.wikipedia.org/wiki/Polar_coordinate_system#Integral_calculus_(area)

(either that or transform to an integral in y and x)

Thanks, I know this formula. :(
Perhaps I am wrong with trying to subtract the line?
If so the solution would be simply:

integral [-pi/4, 3*pi/4] (1/2*((1/(cos(theta)+1))^2)

 

[BvU]

You are not wrong with trying to subtract the line, but it is easier than you seem to think: the line goes though the origin. That is where r=0. So the integration over $dr$ runs from 0 to $r(\theta)$ . And that's easy, because in the relationship:

dA=1/2*r^2 dtheta

the integration over $dr$ (the area of a narrow pie $d\theta$ from 0 to $r(\theta)$) has already been carried out.
So formally you are done according to the problem statement.

I am maniacally checking everything

It would be a very good exercise to also carry out the integration and compare with an integration over $dy$ in the x-y plane

 

[Poetria]

You are not wrong with trying to subtract the line, but it is easier than you seem to think: the line goes though the origin. That is where r=0. So the integration over $dr$ runs from 0 to $r(\theta)$ . And that's easy, because in the relationship:
the integration over $dr$ (the area of a narrow pie $d\theta$ from 0 to $r(\theta)$) has already been carried out.
So formally you are done according to the problem statement.

It would be a very good exercise to also carry out the integration and compare with an integration over $dy$ in the x-y plane

I got:
the equation for the parabola: $x^2+y^2=(1-x)^2$

integral [1-sqrt(2), sqrt(2)+1] (1/2*(1-y^2)+y)
this is 1.8856 (the same result!)

Many thanks. :)

 

[BvU]

You're welcome. Well done !

(can't say I understand the $x^2 + y^2 = 1-x^2$)

 

[Mark44]

I got:
the equation for the parabola: $x^2+y^2={1-x}^2$

What you meant was correct, but it wasn't written correctly in LaTeX. Here's what you surely intended:
$x^2+y^2=(1-x)^2$; i.e., using parentheses around 1 - x, not braces.

If you expand the right side you get $x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1 - 2x$

 

[Poetria]

What you meant was correct, but it wasn't written correctly in LaTeX. Here's what you surely intended:
$x^2+y^2=(1-x)^2$; i.e., using parentheses around 1 - x, not braces.

If you expand the right side you get $x^2 + y^2 = 1 - 2x + x^2 \Rightarrow y^2 = 1 - 2x$

Right. I need more practice in Latex. It was also complicated to write an integral. :(
Many thanks. :) I understand everything now.

Ha I have used wrong brackets {} and not (). :( It is corrected now.

 

[Mark44]

It's not really that complicated, especially since you already know quite a bit of LaTeX -- \int for an indefinite integral, and \int_a^b for definite integral.
Keep in mind that if either integration limit is more than one character, you need braces.
For example $\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta$
In unrendered form: $\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta$

 

[Poetria]

It's not really that complicated, especially since you already know quite a bit of LaTeX -- \int for an indefinite integral, and \int_a^b for definite integral.
Keep in mind that if either integration limit is more than one character, you need braces.
For example $\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta$
In unrendered form: $\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta$

Thank you so much. :) I did it just for practice (well I did have to correct it several times ;) )

$\int_{\theta=-\frac {\pi} 4}^{\frac {3{\pi}} 4} {\frac 1 2} \frac 1 {({\cos(\theta)+1})^2} d\theta$

 

[kent davidge]

In unrendered form: $\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta$

Hei Mark44, how did you manage to post it in unrendered form. Seems like you didn't miss anything so PF should have rendered it because you put it inside ##.

 

[Mark44]

Hei Mark44, how did you manage to post it in unrendered form. Seems like you didn't miss anything so PF should have rendered it because you put it inside ##.

I tricked the browser by making the first # of the start and end tags a different color. That's enough so that the browser doesn't recognize the # pairs.

Here's what the raw text looks like --

In unrendered form: ##\int_{\theta = \pi/4}^{3\pi/4}\cos(\theta)d\theta##

I've done the same trick on the two pairs of BBCode color tags, so that they don't get gobbled up by the browser. BTW, the color I used is black, the same as the other text.

 

[kent davidge]

Oh cool

BTW, the color I used is black, the same as the other text

If you didn't said that, I would think about it.. How did he use different colors if his output is all in black?

 

[BvU]

Dunno, it goes by itself $see$ and $see$

Compare in line math ($)$\int_{\theta=-\frac {\pi} 4}^{\frac {3{\pi}} 4} {\frac 1 2} \frac 1 {({\cos(\theta)+1})^2} d\theta\ ## with displaystyle ($$)
$$\int_{\theta=-\frac {\pi} 4}^{\frac {3{\pi}} 4}\ { \frac 1 2} \frac 1 {({\cos(\theta)+1})^2} \ d\theta$$