# Polar Surface integral

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Thread starter 6 years ago
#1
Question Reads;

Find the integral of the function f(x,y)=x2+y2 over the disc x2+y2 less than or equal to 4.

So I converted this to a polar equation getting

I = r2cos2θ + r2sin2θ r dr dθ

Setting my integration limits to:- r from 0 to 4, and θ from 0 to 2pi

I then seperated the two integrals out, getting

Int(r3) from 4 to 0
and Int (cos2θ + sin2θ) from 2pi to 0

[Dont know how to do an integral sign on this site here, sorry]

Integrating both and then multiplying together I get the overall integral to be: (r4/4)*θ
Using the 2 limits I'd set that gives me the answer to be 64x2pi = 128pi. Which seemed sensible enought o me, howvever when I checked the answer online it gave me something like 256pi/3. For the life of me I cant see how I've gone wrong, so any aid would be appreciated.

Thanks.
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6 years ago
#2
(Original post by Eremor)
...
Your limits on r should be 0 and 2.

I got 8pi.

Where did you check online? If Wolfram, what was your input/URL?
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Thread starter 6 years ago
#3
Ah My bad with the limits.

I've not worked through it with the changed limits yet, But wolfram does give a different answer.

My input was
integration [//math: (r^2)*(cos(q))^2+(r^2)*(Sin(q))^2//]*r [//math:dr dq//] for r from [//math:0//] to [//math:2//]for q from [//math:0//] to [//math:2*pi//]

Remove the space after math: TSR made it a smiley face without it there.

could be doing something wrong, have only just started learning about polar co-ords so its all new to me, Appreciate the help so far.

cheers.
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6 years ago
#4
(Original post by Eremor)
My input was
integration [//math: (r^2)*(cos(q))^2+(r^2)*(Sin(q))^2//]*r [//math:dr dq//] for r from [//math:0//] to [//math:2//]for q from [//math:0//] to [//math:2*pi//]

Remove the space after math: TSR made it a smiley face without it there.

could be doing something wrong, have only just started learning about polar co-ords so its all new to me, Appreciate the help so far.

cheers.
Couldn't work with that, but:

I think your bracketing might have been in error - see here

Note: Cos^2+sin^2=1, and so we can reduce the formula.
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6 years ago
#5
(Original post by Eremor)
Question Reads;

Find the integral of the function f(x,y)=x2+y2 over the disc x2+y2 less than or equal to 4.

So I converted this to a polar equation getting

I = r2cos2θ + r2sin2θ r dr dθ

Setting my integration limits to:- r from 0 to 4, and θ from 0 to 2pi

I then seperated the two integrals out, getting

Int(r3) from 4 to 0
and Int (cos2θ + sin2θ) from 2pi to 0

[Dont know how to do an integral sign on this site here, sorry]

Integrating both and then multiplying together I get the overall integral to be: (r4/4)*θ
Using the 2 limits I'd set that gives me the answer to be 64x2pi = 128pi. Which seemed sensible enought o me, howvever when I checked the answer online it gave me something like 256pi/3. For the life of me I cant see how I've gone wrong, so any aid would be appreciated.

Thanks.
ghostwalker is correct

look at some more questions on the link

http://madasmaths.com/archive_maths_...ed_topics.html

download file

multivariable_integration_in_polars
(Be patient long download time)

Good luck
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