areacalculuspolar coordinates
I am trying to find the shaded area and this is how I am doing it.
$$\frac12\int_{2\pi}^{3\pi} \theta^2 \, d\theta = \frac 1 2 \left[{\frac{\theta^3}3}\right]_{2\pi}^{3\pi} = \frac12\left(\frac{27\pi^3}{3}-\frac{8\pi^3}{3}\right)$$
$$My\,Calculation\; \frac{19\pi^3}{6} $$
$$Expected\,Answer\;3\pi^3$$
Is the expected answer wrong? or Am I doing something wrong?
First, to sketch such a graph, you want to consider the distance from the origin as the angle from the $x$-axis changes. Just like when sketching the graph of a function in rectangular coordinates it is good to evaluate at particular values of $x$ and see the height of the function, when sketching a curve in polar coordinates, evaluate the function are a few values of the angle and find the radius, i.e., the distance from the origin at the angle. So, doing that we obtain the following graph:
Then, to calculate the area of the radial set, you must integrate $\frac{1}{2} r^2$, where the radius is the value of the function. So we have, \begin{align*} \text{Area} &= \frac{1}{2}\int_0^{2 \pi} \theta^2 \, d\theta \\ &= \left. \frac{1}{2} \cdot \frac{\theta^3}{3} \right|_0^{2 \pi} \\ &= \frac{(2 \pi)^3}{6}\\ &= \frac{4 \pi^3}{3}. \end{align*}
Note that the integral limits for the circle is $[-\frac\pi2, \frac\pi2]$. Thus, the integral should be set up as
$$\frac12\int^{\pi}_{-\pi}\left(1+\cos\theta\right)^2d\theta -\frac12\int^{\pi/2}_{-\pi/2}\cos^2\theta\ d\theta = \frac{5\pi}4$$
Best Answer
That integral gives the area bounded by that section, but totally ignores the area cut away by a previous section of the curve, specifically that on $[0, \pi]$. Computing and subtracting that gives the desired result.