Recall that the area of a region $R$ in the $xy$ plane is $A_{xy}=\ds\iint_{R_{r\theta}} r dr d\theta$ where $R_{r\theta}$ is the region in the polar plane.
- Draw the region in the $xy$ plane whose area is given by $\int_0^{3\pi/2}\int_1^{4-3\cos\theta} r dr d\theta$. I'll provide the graph of $r=4-3\cos\theta$ to get you started.
- To find the area inside one petal of the rose $r=3\sin 2\theta$, Joe tries using $A = \int_0^{\pi/2}\int_0^{3}rdrd\theta.$ Draw the region described by these bounds, and then correct his mistake.
- To find the area inside the cardiod $r=3-3\sin\theta$, Jill tries using $A = \int_0^{2\pi}\int_0^{3-3\sin\theta}(3-3\sin\theta)drd\theta.$ Why is her solution incorrect? Find and fix the mistake.
- Find the area of the region cut from the first quadrant by the cardioid $r=1+\sin \theta$
- Find the area inside one leaf of the rose $r=\cos 3\theta$
- Find the area inside the cardioid $r=1+\cos \theta$ and outside the curve $r=1$.
- Change the following Cartesian integrals into polar integrals (Hint: draw the region to determine the bounds)
- $\int_{-1}^1 \int_0^{\sqrt{1-x^2}} dy \, dx$
- $\int_{0}^2 \int_0^{\sqrt{4-y^2}} \, (x^2 + y^2) \, dx \, dy$
- $\int_{0}^6 \int_0^y \, x \, dx \, dy$
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