Finding the area of the region defined in polar coordinates by $0\leq\theta\leq\pi$ and $0\leq r\leq\theta^3$

Question

Find the area of the region defined in polar coordinates by $0 \leq \theta \leq \pi$ and $0 \leq r \leq {\theta}^3$.

I tried using the formula

$$A = \int_{0}^{\pi} \frac{1}{2} r(\theta)^2 d\theta$$

However, I don't know which value I should use for $r(\theta)$. I tried using ${\theta}^3$ but it marked it wrong. Here is my work:

$$A = \frac{1}{2} \int_{0}^{\pi} {\theta}^3 d\theta$$

$$A = \left(\frac{1}{2}\right) \left(\frac{1}{4}\right) \theta^4 \Big\rvert_{0}^{\pi}$$

$$A = \frac{1}{8}(\pi)^4$$

What value should I use for $r(\theta)$? What am I missing?

Answer 1

$$\large A=\int_0^\pi\int_0^{\theta^3}rdrd\theta=\int_0^\pi\frac12r^2\Big|_0^{\theta^3}d\theta=\int_0^\pi\frac12\theta^\color{red}6d\theta$$

Answer 2

The area is $$\int \int rdrd\theta=$$

$$\int_0^{\pi}(\int_0^{\theta^3}rdr)d\theta=$$

$$\int_0^{\pi}\Bigl[\frac{r^2}{2}\Bigr]_0^{\theta^3}d\theta=$$

$$\frac 12\int_0^{\pi}\theta^6d\theta=$$

$$\frac{\pi^7}{14}$$