# Polar Surface integral

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Question Reads;

Find the integral of the function f(x,y)=x

So I converted this to a polar equation getting

I = r

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(r

and Int (cos

Integrating both and then multiplying together I get the overall integral to be: (r

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.

Find the integral of the function f(x,y)=x

^{2}+y^{2 }over the disc x^{2}+y2 less than or equal to 4.So I converted this to a polar equation getting

I = r

^{2}cos^{2}θ + r^{2}sin^{2}θ r dr dθSetting my integration limits to:- r from 0 to 4, and θ from 0 to 2pi

I then seperated the two integrals out, getting

Int(r

^{3}) from 4 to 0and Int (cos

^{2}θ + sin^{2}θ) 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: (r

^{4}/4)*θUsing 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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#2

(Original post by

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**Eremor**)...

I got 8pi.

Where did you check online? If Wolfram, what was your input/URL?

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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//]

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.

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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#4

(Original post by

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//]

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.

**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.

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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#5

(Original post by

Question Reads;

Find the integral of the function f(x,y)=x

So I converted this to a polar equation getting

I = r

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(r

and Int (cos

Integrating both and then multiplying together I get the overall integral to be: (r

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.

**Eremor**)Question Reads;

Find the integral of the function f(x,y)=x

^{2}+y^{2 }over the disc x^{2}+y2 less than or equal to 4.So I converted this to a polar equation getting

I = r

^{2}cos^{2}θ + r^{2}sin^{2}θ r dr dθSetting my integration limits to:- r from 0 to 4, and θ from 0 to 2pi

I then seperated the two integrals out, getting

Int(r

^{3}) from 4 to 0and Int (cos

^{2}θ + sin^{2}θ) 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: (r

^{4}/4)*θUsing 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.

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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