# double integrals in polar coordinates

• December 8th 2010, 09:26 PM
danyellogram
double integrals in polar coordinates
I have to use a double integral in polar coordinates to find the area of the region inside the circle x^2 + y^2 = 4 and to the right of the line x =1.

I have it all graphed and I know how to do double integrals but I am confused as to what I am integrating, it is probably really simple but I was sick the day we went over this section in class so I have no idea what I am doing.

• December 8th 2010, 10:33 PM
Prove It
Start by writing your double integral in cartesians first.

First, where do the two graphs intersect? When $\displaystyle x = 1$.

So $\displaystyle 1^2 + y^2 = 4$

$\displaystyle 1 + y^2 = 4$

$\displaystyle y^2 = 3$

$\displaystyle y = \pm \sqrt{3}$.

So if you take horiztonal strips, your $\displaystyle x$ values range from $\displaystyle 1$ to the graph $\displaystyle x = \sqrt{4 - y^2}$, and your $\displaystyle y$ values range from $\displaystyle -\sqrt{3}$ to $\displaystyle \sqrt{3}$.

So what is your double integral in cartesians?
• December 9th 2010, 10:24 AM
danyellogram
double integrals in cartesian coordinates
Here's a picture of what I have so far, I hope I am going in the right direction:
Attachment 20042

would i replace x^2 + y^2 with r? if so what would my boundaries be? I'm confused by this....
• December 9th 2010, 11:41 AM
AllanCuz
Quote:

Originally Posted by danyellogram
I have to use a double integral in polar coordinates to find the area of the region inside the circle x^2 + y^2 = 4 and to the right of the line x =1.

I have it all graphed and I know how to do double integrals but I am confused as to what I am integrating, it is probably really simple but I was sick the day we went over this section in class so I have no idea what I am doing.

For future problems you can reference my tutorial on double integration at the top of this forum.

I'll offer a different approach to Prove It (actually it's the same approach just worded differently) and outline my approach via steps so that you can follow the same process for other questions. Though I wouldn't write this in cartesian coordinates before moving to polar, I would go straight to polar.

Quote:

Step 1: Draw the Domain
In this case we have a circle of radius 2 and a vertical line at x = 1. If we want the area to the right of this then we are looking at two quarter circles. For simplicity we will compute the area of 1 of these quarters and multiply by 2.

Quote:

Step 2: Set up the integral in general terms
$\iint_D f(x) dA$

Note that before we can approach this problem we need to find a set of bounds. Well, these bounds depend on which way we want to integrate our function.

This actually has some complexity to it because we are computing this integral in polar co-ordinates.

Quote:

Step 3: Find Bounds
Let's look at the area of a regular circle of radius 2.

$A = \int_0^{2 \pi } d \theta \int_0^2 rdr = \pi (2^2)$

But in our case we have a nice line going right up the middle and our radius is no longer constant. If you draw a line from the origin to the radius of the circle you'll see that it crosses the line x=1. Well, we dont want anything before that line so if we make a triangle we see that

$cos \theta = \frac{ 1 }{R} \to R= \frac{1}{cos \theta}$

Well this is good, so we now have our bounds for R

$\frac{1}{cos \theta } \le R \le 2$

We now need to evaluate theta. Well clearly we start at 0 (defined from the x-axis) but where do we end? Well we end where the line x=1 intersects the circle. This happens when x=1 so when we sub x=1 into our circle, we get the y bound of $y = \sqrt{3}$. Now to get the theta angle we have a triangle with a height of square root 3 and a width of 1.

$Tan \theta = \frac{ \sqrt{3} }{1} \to \theta = 60$

So theta runs from 0 to 60.

Quote:

Step 4: Put it all together
$A = 2 \int_0^{ \frac{ \pi }{3} } d \theta \int_{ \frac{1}{cos \theta} }^2 r dr$
• December 9th 2010, 02:44 PM
danyellogram
thank you both for your help! <3
• December 9th 2010, 03:50 PM
Prove It
Quote:

Originally Posted by danyellogram
Here's a picture of what I have so far, I hope I am going in the right direction:
Attachment 20042

would i replace x^2 + y^2 with r? if so what would my boundaries be? I'm confused by this....

I also realise I have made a mistake which you have taken through to your working - the $\displaystyle x$ values range from $\displaystyle +1$ to $\displaystyle \sqrt{4 - y^2}$.

Also, if you are taking horizontal strips, then you are integrating with respect to $\displaystyle x$ first. So that means the double integral should be

$\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}{\int_{1}^{\sqrt{4-y^2}}{1\,dx}\,dy}$. Now follow Allan's steps to convert to polars.

Epilogue: You don't even need to convert to polars:

$\displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}{\int_{1}^{\sqrt{4-y^2}}{\,dx}\,dy} = \int_{-\sqrt{3}}^{\sqrt{3}}{\left[x\right]_1^{\sqrt{4-y^2}}\,dy}$

$\displaystyle = \int_{-\sqrt{3}}^{\sqrt{3}}{\sqrt{4 - y^2} - 1\,dy}$

$\displaystyle = \int_{-\sqrt{3}}^{\sqrt{3}}{\sqrt{4 - y^2}\,dy} - \int_{-\sqrt{3}}^{\sqrt{3}}{1\,dy}$

Now you can make a trigonometric substitution to solve.
• December 9th 2010, 06:25 PM
danyellogram
Yes when I was working it out (before allan's post) I found that slight mistake :D

again, thank you for all of your help.