- To find the area inside one petal of the rose $r=3\sin 2\theta$, one person tries using $A = \int_0^{\pi/2}\int_0^{3}rdrd\theta.$ Draw the region described by their bounds, and then correct their mistake.
- To find the area inside the cardiod $r=3-3\sin\theta$, one person tries using $A = \int_0^{2\pi}\int_0^{3-3\sin\theta}(3-3\sin\theta)drd\theta.$ Why is their solution incorrect? Find and fix their mistake.
- We have seen from the product rule that
$$ \begin{pmatrix}dx\\dy\end{pmatrix}= \begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}dr+ \begin{pmatrix}-r\sin\theta\\r\cos\theta\end{pmatrix}d\theta.$$
- Let $r=2,\theta=\pi/3,dr=1,d\theta=0$. Simplify $(dx,dy)$ and then add this vector with its base at $(r,\theta)=(2,\pi/3)$ to the picture on the board.
- Let $r=2,\theta=\pi/3,dr=0,d\theta=\pi/6$. Simplify $(dx,dy)$ and then add this vector with its base at $(r,\theta)=(2,\pi/3)$ to the picture on the board.
- The two vectors above form the edges of a parallelogram. Shade this parallelogram and find its area.
- If instead for the first vector we change to $dr=.2$, and for the second we change to $d\theta=\pi/24$, shade the corresponding paralellogram and find its area.
- If instead for the first vector we change to $dr=dr$, and for the second we change to $d\theta=d\theta$, find its area.
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