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Use a double integral to find the area enclosed by one loop of the four-
leaved rose $$r=\cos 2 \theta .$$

Area $$=\int_{y} \int_{x} d x d y=\int_{r_{1}}^{r_{2}} \int_{\theta_{1}}^{\theta_{2}} r d r d \theta$$

$$\int_{-{\pi/4}}^{\pi/4} \int_{0}^{\cos 2 \theta} r d r d \theta$$

$$=\int_{-\frac{\pi}{4}}^\frac {\pi}{4}\left[\int_{0}^{\cos 2 \theta} r d r\right] d \theta$$

$$=\int_{-\frac{\pi}{4}}^{\frac {\pi}{4}} \int_{0}^{\cos 2 \theta} \frac{r^{2}}{2} d\theta$$

$$=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos (2 \theta)^{2}}{2}=\frac{\pi}{8} $$

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