Area between a circle and an spiral.

Question

I have to evaluate the following area:

Since the area of the circle is $16\pi$, I thought about evaluate the area of the spiral from $0$ to $4$: $$\frac{1}{2}\int_{0}^{4}\theta^2 d\theta=\frac{32}{3}$$ because the curves have intersection when $r=\theta=4$. And since the area of the 4th quadrant is $4\pi$ then, my final answer is $$16\pi-\frac{32}{3}-4\pi=12\pi-\frac{32}{3}$$

I'd lke to know if it's correct.

Answer 1

Back to basics: The area in polar coordinates is given by

$$ A=\frac{1}{2}\int r^2d\theta $$

So, here we have

$$ \begin{align} A&=\frac{1}{2}\int_0^4 (r^2-\theta^2)d\theta\\ &=\frac{1}{2}\bigg[r^2\theta-\frac{\theta^3}{3} \bigg]_0^4\\ &=\frac{1}{2}\bigg[4^3-\frac{4^3}{3} \bigg]\\ &=\frac{64}{3}\approx 21.3333 \end{align} $$

I've verified this result numerically in the complex plane.

Answer 2

Taking inspiration from the area under a cycloid

Consider the non-self-intersecting plane curve defined by the parametric equations $$x=(t), y=y(t), a\leq t \leq b$$ assume $x(t)$ differentiable, the area between the curve and $x$-axis is $$A = \int_a^by(t)x'(t)dt$$

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Your spiral has parametric equation $$\gamma(t) = (t\cos (t), t\sin(t)), \ t\in [0,4] $$ And you can divide in three section

1. The area between $y$-axis and curve from $0$ to $\pi/2$ (this interval is on the $y$-axis)
2. Area between curve and $x$-axis from $0$ to $-\pi$
3. Last part, area between the curve and $y$-axis from $0$ to $4\sin(4)$

With the formula above you should be able to evaluate the portions mentioned. I honestly tried to evaluate only the part first part (using the formula with respect to $y$-axis) $$\int_0^{\pi/2}x(t)y'(t)dt =\int_0^{\pi/2}\big(t\cos(t)\sin(t) + t^2 \cos^2(t)\big)dt= \pi^3/48 $$ I think is not that bad, maybe this isn't the most efficient way. Hope this helps!