Math 215 - Multivariate Calculus - Schedule

Draw a bear in polar coordinates, and then Cartesian coordinates.

If you are a little witty over a period of time, what do you get?

Solution

You get totally witty. $$\int_{time}d(\text{wit}) = \text{total wit} $$

Rapid Recall

Find the area of a parallelogram whose edges are parallel to the vectors $(\cos\theta dr,\sin\theta dr)$ and $(-r\sin\theta d\theta,r\cos\theta d\theta)$.
When we add up lots of little bits of area, so $\int_R dA$, what do we get?
Draw the region described by $\pi/6\leq \theta\leq \pi/3$ and $4\leq r\leq 5$.
For the curve $r=7$, compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.

Solutions

$|r\cos^2\theta dr d\theta +r\sin^2\theta dr d\theta| = |r dr d\theta|$
Total Area
See board
Since $x=r\cos\theta = 7\cos\theta$ and $y=r\sin\theta=7\sin\theta$, we have $\frac{dx}{d\theta} = -7\sin\theta$ and $\frac{dy}{d\theta}=7\cos\theta$.

Consider the polar curve $r=7$. Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$. Then use the arc length formula $$\int_C\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$$ to find the arc length for the portion of this curve with $0\leq \theta\leq \alpha$.
Draw the region in the plane described by $-3\leq y\leq 2$ and $y\leq x\leq 6-y^2$.
Compute the integral $\ds\int_{y}^{6-y^2}dx$ (assume $y$ is a constant).
Compute the double integral $\ds \int_{-3}^{2}\left(\int_{y}^{6-y^2}dx\right)dy$.
Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi$ and $2\leq r\leq 5$.
Compute the double integral $\ds \int_{0}^{\pi}\left(\int_{2}^{5}rdr\right)d\theta$.
Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/3$ and $0\leq r\leq 2\sin3\theta$.
Draw the region in the $xy$ plane described by $0\leq \theta \leq \pi/4$ and $0\leq r\leq 3\sin2\theta$.
Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the cardiod $r=2+2\cos\theta$.
Set up a double integral that gives the area of the region in the $xy$ plane that lies inside one petal of the rose $r=3\cos2\theta$.
Draw and shade the region in the $xy$ plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.
Set up a double integral that gives the area of the region in the $xy$ plane that lies inside the curve $r=3+2\cos\theta$ and outside the curve $r=1$.