Area in polar coordinate using multiple integral

[ahmed markhoos]

The question is to find the area of a disk, r ≤ 2a×cos(θ)
as in the figure "example-just an illustration"

I used two methods, each gave different wrong answers

- integrate 2a×cos(θ) dθ dr - from θ=0 to θ=π/2 and from r=2a to r=2a×cos(θ) ; then I simply multiplied the answer by 2.

- integrate 2a×cos(θ) dθ dr - from θ=0 to θ=2π and from r=2a to r=2a×cos(θ)

I was hopping to get the right answer using the first method, but surprisingly it didn't work, integration seems to be fine I even used wolfram to check it.
-----------------

I have another small question. I had a question about finding the volume of flipped cone "its point in the origin and its shape is in the positive side of z-axis", its height is the same as its base radius so that, r=h z=h, thus r=z
"example-just an illustration"

http://01.edu-cdn.com/files/static/mcgrawhillprof/9780071624756/INVERTED_CONE_WATER_TANK_PROBLEM_01.GIF

thing is, when I put the limits say : r=0 ⇒ h //// z=0 ⇒ r //// θ=0 ⇒ 2π . It give wrong answer

but r=0 ⇒ z //// z=0 ⇒ h //// θ=0 ⇒ 2π . give the right one

that's really confusing!

 

[Ray Vickson]

The question is to find the area of a disk, r ≤ 2a×cos(θ)
as in the figure "example-just an illustration"

I used two methods, each gave different wrong answers

- integrate 2a×cos(θ) dθ dr - from θ=0 to θ=π/2 and from r=2a to r=2a×cos(θ) ; then I simply multiplied the answer by 2.

- integrate 2a×cos(θ) dθ dr - from θ=0 to θ=2π and from r=2a to r=2a×cos(θ)

I was hopping to get the right answer using the first method, but surprisingly it didn't work, integration seems to be fine I even used wolfram to check it.I have another small question. I had a question about finding the volume of flipped cone "its point in the origin and its shape is in the positive side of z-axis", its height is the same as its base radius so that, r=h z=h, thus r=z
"example-just an illustration"

http://01.edu-cdn.com/files/static/mcgrawhillprof/9780071624756/INVERTED_CONE_WATER_TANK_PROBLEM_01.GIF

thing is, when I put the limits say : r=0 ⇒ h //// z=0 ⇒ r //// θ=0 ⇒ 2π . It give wrong answer

but r=0 ⇒ z //// z=0 ⇒ h //// θ=0 ⇒ 2π . give the right one

that's really confusing!

In polar coordinates the area element is $dA = r \, dr \, d \theta$. To get the area inside the curve $r = 2a \cos(\theta)$ we need to integrate $dA$ for $r =0 \to 2 a \cos(\theta)$ and for $\theta = 0 \to \pi/2$, then multiply the result by 2 (or integrate for $\theta = -\pi/2 \to \pi/2$).

The boundary curve $r = 2a \cos(\theta)$ has a simple form when expressed in cartesian coordinates---looking at the plot suggests what it should be. From the resulting form one can easily determine the interior area with essentially no work.

 

[ahmed markhoos]

In polar coordinates the area element is $dA = r \, dr \, d \theta$. To get the area inside the curve $r = 2a \cos(\theta)$ we need to integrate $dA$ for $r =0 \to 2 a \cos(\theta)$ and for $\theta = 0 \to \pi/2$, then multiply the result by 2 (or integrate for $\theta = -\pi/2 \to \pi/2$).

The boundary curve $r = 2a \cos(\theta)$ has a simple form when expressed in cartesian coordinates---looking at the plot suggests what it should be. From the resulting form one can easily determine the interior area with essentially no work.

we begin at $θ=0$, why should $r=0$ ?, logicaly $r=2acosθ$ give us $r=2a$ !
the thing is, $r$ is the distance between the point and the origin and $θ$ is the angle between r and x-axis !

I don't know how did you do it but! the limits you suggest were wrong "both limit from $-π/2$ to $π/2$ and $0$ to $π/2$ multiplied by $2$" give an answer of$= 2πa^2$, which is double the area we expect.

Thank you.

 

[Ray Vickson]

we begin at $θ=0$, why should $r=0$ ?, logicaly $r=2acosθ$ give us $r=2a$ !
the thing is, $r$ is the distance between the point and the origin and $θ$ is the angle between r and x-axis !

I don't know how did you do it but! the limits you suggest were wrong "both limit from $-π/2$ to $π/2$ and $0$ to $π/2$ multiplied by $2$" give an answer of$= 2πa^2$, which is double the area we expect.

Thank you.

Look at the picture! For a given value of $\theta$, $r$ starts at $r = 0$ and ends at $r = 2a \cos(\theta)$.

My limits are not wrong, and when I do the integration I get an area of exactly $2 \times \pi a^2/2 = \pi a^2$ in the $0 \to \pi/2$ case and $\pi a^2$ directly in the $-\pi/2 \to \pi/2$ case. Yes, I have done the integrations, and that is what I get!

Please re-read what I said in post #2. There I told you exactly what integrations I performed, and they are not at all the same as the ones you did.

 

[ahmed markhoos]

but the computer gave me solutions different than yours
look

http://im50.gulfup.com/rtRHjt.png

http://im50.gulfup.com/lL8Idy.png

 

[Ray Vickson]

but the computer gave me solutions different than yours
look

http://im50.gulfup.com/rtRHjt.png

http://im50.gulfup.com/lL8Idy.png

Those are not the correct integrals; they do not represent the area you want. Again: re-read (carefully!) what I wrote; or else, look up the topic in a book or via Google.

 

[ahmed markhoos]

Those are not the correct integrals; they do not represent the area you want. Again: re-read (carefully!) what I wrote; or else, look up the topic in a book or via Google.

Got it!, thank you. But can I ask another question ?

why can't I substitute $r=2a cos(θ)$ in the integral?, I did it because it seems reasonable in an integral. I mean since r depend on θ, then I can't integrate $2a cos(θ)$ which is in $r$ with respect to r because it contain a constant!. All that doesn't matter, still the question is why I can't substitute with the value of r ?

is it because the disk is represented by $r ≤ 2a cos(θ)$ ??

 

[Ray Vickson]

Got it!, thank you. But can I ask another question ?

why can't I substitute $r=2a cos(θ)$ in the integral?, I did it because it seems reasonable in an integral. I mean since r depend on θ, then I can't integrate $2a cos(θ)$ which is in $r$ with respect to r because it contain a constant!. All that doesn't matter, still the question is why I can't substitute with the value of r ?

is it because the disk is represented by $r ≤ 2a cos(θ)$ ??

Basically, yes.

I did suggest you look up the topic in Google. Have you done that? You should; there are several nice articles with derivations, diagrams, worked examples, etc. All your questions will be answered there.