Polar coordinates area of the region bounded problem

**muddyjch** · July 21st 2010, 10:08 AM

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.

**HallsofIvy** · July 21st 2010, 12:01 PM

Originally Posted by muddyjch

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** · July 21st 2010, 12:20 PM

Hello, muddyjch!

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

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

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

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

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

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

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

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

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

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

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

. . . . . . . . . . $=\;89\pi \;\approx\;279.6017462$

But check my work . . . please!

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