Polar coordinates area of the region bounded problem

Jul 2010
4
1
Find the area of the region bounded by: r = 9 − 4 sinθ

I tried this multiple times and the answer i got was 381.7035 but that is not correct.
 
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HallsofIvy

MHF Helper
Apr 2005
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7,909
Find the area of the region bounded by: r = 9 − 4 sinθ

I tried this multiple times and the answer i got was 381.7035 but that is not correct.
It looks straightforward to me (and I do NOT get 381.7035). What exactly did you do?
 

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, muddyjch!

Find the area of the region bounded by: .\(\displaystyle r \:=\: 9 - 4\sin\theta\)

This is a heart-shaped region symmtetric to the "y-axis".

We can integrate from \(\displaystyle \text{-}\frac{\pi}{2}\) to \(\displaystyle +\frac{\pi}{2}\) and multiply by 2.


We have: .\(\displaystyle \displaystyle{A \;=\;2 \times \frac{1}{2}\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} (9 - 4\sin\theta)^2\,d\theta} \)

. . . . . . . . . .\(\displaystyle \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} \left(81 - 72\sin\theta + 16\sin^2\!\theta\right)\,d\theta }\)

. . . . . . . . . .\(\displaystyle \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(81 - 72\sin\theta + 16\left[\tfrac{1-\cos2\theta}{2}\right]\bigg)\,d\theta }\)

. . . . . . . . . .\(\displaystyle \displaystyle{=\; \int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(81 - 72\sin\theta + 8 - 8\cos2\theta\bigg)\,d\theta }\)

. . . . . . . . . .\(\displaystyle \displaystyle{=\;\int^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}}\bigg(89 - 72\sin\theta - 8\cos2\theta\bigg)\,d\theta }\)

. . . . . . . . . .\(\displaystyle \displaystyle{=\;89\theta + 72\cos\theta - 4\sin2\theta\,\bigg]^{\frac{\pi}{2}}_{\text{-}\frac{\pi}{2}} }\)

. . . . . . . . . .\(\displaystyle =\; \bigg[89\left(\frac{\pi}{2}\right) + \cos\frac{\pi}{2} - 4\sin\pi\bigg] - \bigg[89\left(\text{-}\frac{\pi}{2}\right) + 72\cos\left(\text{-}\frac{\pi}{2}\right) - 4\sin(\text{-}\pi)\bigg] \)

. . . . . . . . . .\(\displaystyle =\;\bigg(\frac{89\pi}{2} + 0 - 0\bigg) - \bigg(\text{-}\frac{89\pi}{2} + 0 - 0\bigg)\)

. . . . . . . . . .\(\displaystyle =\;89\pi \;\approx\;279.6017462 \)


But check my work . . . please!