# Polar Graph - Area between curves

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Thread starter 1 month ago
#1
https://imgur.com/a/CRc5yyH
Please find question in the above link.

I found point of intersection at theta = pi/4

Integrated 0.5(1+sin theta)^2 to find area from theta = 0 to pi/4. I got 0.5(3pi/8- sqrt 2 + 7/4)

Integrated 0.5(1+cos theta)^2 to find area from theta = pi/4 to pi. I got 0.5(9pi/8 - sqrt 2 - 1/4)

Added and got 3pi/4 - sqrt 2 + 3/4

Answer is in fact 3pi/4 + sqrt 2 + 5/4.

0
1 month ago
#2
(Original post by golgiapparatus31)
https://imgur.com/a/CRc5yyH
Please find question in the above link.

I found point of intersection at theta = pi/4

Integrated 0.5(1+sin theta)^2 to find area from theta = 0 to pi/4. I got 0.5(3pi/8- sqrt 2 + 7/4)

Integrated 0.5(1+cos theta)^2 to find area from theta = pi/4 to pi. I got 0.5(9pi/8 - sqrt 2 - 1/4)

Added and got 3pi/4 - sqrt 2 + 3/4

Answer is in fact 3pi/4 + sqrt 2 + 5/4.

Your actual integrations are correct, but I can't tell if you're integrating the right functions or with the correct limits as my machine can't handle imgur.com any more.
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Thread starter 1 month ago
#3
(Original post by ghostwalker)
Your actual integrations are correct, but I can't tell if you're integrating the right functions or with the correct limits as my machine can't handle imgur.com any more.

I have tried to attach it in this post:
1
1 month ago
#4
(Original post by golgiapparatus31)

I have tried to attach it in this post:
Yep, got that.

In which case your two integrals are correct.

Using some software, I can confirm that you are correct, and their answer is just way to big.
Last edited by ghostwalker; 1 month ago
1
Thread starter 1 month ago
#5
(Original post by ghostwalker)
Yep, got that.

In which case your two integrals are correct.

Using some software, I can confirm that you are correct, and their answer is just way to big.
Thanks so much
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