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I am having some problem with caluculating areas using double integrals in polar coordinates. The question is : Calculate the total area of the rose $ r = 5 \sin(2 \theta)$ using double integration in polar coordinates?

I took the limits of theta from $0$ to $\pi \over 4$ and that for $r$ from $0$ to $5 \sin(2 \theta)$ and multiplied it by $2$ to get the area of one petal of the rose and then multiplied it by 4 to get the total area.

My answer comes out to be $ 25 \pi \over 2$. Have I followed the right procedure and is the answer correct?

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1 Answer 1

$$ \text{Area} = \int_0^{2 \pi} \int_0^r r dr\; d\theta= 4\int_0^{\pi \over 2} {1 \over 2}r^2 d\theta = 4\int_0^{\pi \over 2} {1 \over 2}25 \sin^2 2\theta d\theta \\ 50 \int_0^{\pi \over 2}{1 - \cos 4 \theta \over 2} d\theta = 25\left[\theta - {\sin 4\theta \over 4} \right]_0^{\pi \over 2} = {25 \pi\over 2}$$

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Seems so!!

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