Find the area below the polar curve $r=16+16\sin(\theta)$ and above line $y=8$

Question

So, this questioned popped up on a homework in my Calculus II class, and I'm pretty confused. I am familiar with finding the area between two polar curves or two "Cartesian" functions but not a mixture of the two. I've plotted the two functions together so I have a sense of what area I'm trying to find.

The general formula for the area of a polar curve is $\int_a^b\frac12(r)^2d\theta$ where $r(\theta)$ is our function, and $a$ and $b$ are values of $\theta$. If I were tasked with finding the area between two polar curves, I would subtract from the area of the "outer" curve the area of the "inner" curve. I also know that for regular functions of x or y, a similar procedure applies.

But how do I use this here? How do I identify the bounds of integration? And once I do, what do I integrate? I've tried several things and none of it seems to work. Any help wrapping my head around this would be much appreciated.

Answer 1

Find where these functions equal: $$r=16(1+\sin(\theta))$$ $$y=8$$ $$r\sin(\theta)=8$$ Substitute into the first equation: $$\frac{8}{\sin(\theta)}=16(1+\sin(\theta))$$ $$\sin^2(\theta)+\sin(\theta)-\frac{1}{2}=0$$

Use quadratic formula: $$\sin(\theta)=\frac{-1\pm\sqrt{3}}{2}$$ $$\theta_{2}=\arcsin(\frac{-1+\sqrt{3}}{2})$$ $$\theta_{1}=-\arcsin(\frac{-1+\sqrt{3}}{2})$$ We can now apply the Area Between Two Curves Formula:

$$\int_{\theta_{1}}^{\theta_{2}}|16(1+\sin(\theta))-\frac{8}{\sin(\theta)}|d\theta=|\int_{\theta_{1}}^{\theta_{2}}16(1+\sin(\theta))-\frac{8}{\sin(\theta)}d\theta|$$ $$|\int_{\theta_{1}}^{\theta_{2}}16d\theta+16\int_{\theta_{1}}^{\theta_{2}}\sin(\theta)d\theta-8\int_{\theta_{1}}^{\theta_{2}}\csc(\theta)d\theta|$$ $$|16(\theta_{2}-\theta_{1})-16\cos(\theta)|_{\theta_{1}}^{\theta_{2}}+8\ln(|\csc(\theta)+\cot(\theta)|)|_{\theta_{1}}^{\theta_{2}}|$$

Now using some trig identities you can evaluate the values for $\theta$ and get your answer.

EDIT:My orignial answer for $\theta_{1}=\arcsin(\frac{-1-\sqrt{3}}{2})$ has a domain error. However it is evident from the graph of this function that the roots are just reflections across the y-axis.

Answer 2

First let's study when do they intersect.

$$r = 16 + 16 \sin \theta$$

$$y=8$$

multiply $\sin \theta$ to the first equation.

$$8= 16\sin \theta + 16 \sin^2 \theta$$

$$2 \sin^2 \theta + 2\sin\theta-1 = 0$$ This is a quadratic equation, I will leave this as an exercise, let the smallest positive solution be $\theta_0$, of which from there we can compute the corresponding $r_0$.

The intersection point is $(r_0 \cos \theta_0, r_1 \sin \theta_1)$.

Notice that the region of interest is symmetrical about the $y$-axis, hence we can focus on the region on the right hand side.

If we connect the intersection point to the origin. This line, $y=8$, and the $y$-axis form a triangle, of which the area should be computable, I shall call this $A_\Delta$.

Hence the quantity of interest is

$$2\left(\int_{\theta_0}^{\frac{\pi}{2}} \int_0^{26+16\sin \theta} r\,\, dr d\theta - A_\Delta\right)$$

Answer 3

Alternatively, the polar curve can be represented parametrically since $(x, y) = (r \cos \theta, r \sin \theta)$, where we let $r$ be a function of $\theta$.

Let us then scale all the lengths by a factor of $8$, so that $r = 2 + 2 \sin \theta$, $y = 1$, and the new area is thus $8^2 = 64$ times smaller. Then we have $(x(t), y(t)) = ((2 + 2 \sin \theta)\cos \theta, (2 + 2 \sin \theta) \sin \theta)$. Finding the values of $\theta$ such that $y = 1$ (we know from the graph that there are exactly two):

$$2 \sin^2 \theta + 2 \sin \theta - 1 = 0$$ $$\sin \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 2 \cdot -1}}{4} = \frac{-1 \pm \sqrt{3}}{2}$$

The given area is just the area swept out between these two values of $\theta$ minus the area of a triangle. The first value of $\theta$ is $\arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)$, but the other value of $\sin \theta$ does not lie between $-1$ and $1$. Considering the symmetry of the graph across the $y$-axis however, $r$ obtains a maximum at $x = \pi/2$, and so $\theta_2 = \arcsin \left(\frac{-1 + \sqrt{3}}{2}\right) + 2 \left(\frac{\pi}{2} - \arcsin \left(\frac{-1 + \sqrt{3}}{2} \right)\right) = \pi - \arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)$.

Thus we obtain the area as:

$$A_1 = \int_{\arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)}^{\pi - \arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)} y(\theta) \ \frac{dx}{d \theta} \ d \theta$$ $$ = \int_{\arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)}^{\pi - \arcsin \left(\frac{-1 + \sqrt{3}}{2}\right)} (2 \sin^2 \theta + 2 \sin \theta)(-2 \sin \theta + 2 \cos(2 \theta)) \ d \theta.$$

$$A_2 = \frac{1}{2} \cdot 1 \cdot 2 \left(\left(2 + 2\left(\frac{-1 + \sqrt{3}}{2}\right) \right) \left(\sqrt{1 - \left(\frac{-1 + \sqrt{3}}{2}\right)^2} \right)\right)$$

and $A_\text{total} = A_1 + A_2$, which we then multiply by $64$ to get the final answer.