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

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.

• Note your integral should be $d\theta$, not $dr$. At what values of $\theta$ do the two curves intersect? (And, first, what is the equation for $y=8$ in polar coordinates?) – Ted Shifrin Nov 14 '17 at 1:51
• Oops, just edited the integral. Thank you! At first I thought I could just set $16+16\sin\theta=8$, but now I see that doesn't make sense, and that I'd need the polar equation for $y=8$. If $y=r\sin\theta$, then $r=\frac{y}{\sin\theta}$. Substituting $y=8$, we get $r=\frac{8}{\sin \theta}$. Do I then set this equal to $16+16\sin\theta$? – akot717 Nov 14 '17 at 2:00

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.

• I should probably know the closed forms for the $\arcsin$ expressions but I'm kinda blanking on them right now. – aleden Nov 14 '17 at 2:16
• Thank you!! Makes sense now. I didn't think to use the quadratic formula for some reason. – akot717 Nov 14 '17 at 2:16
• When I try to compute this using a program like Mathematica, it gives me the following answer: $29.5419 + 20.7236i$. How do I retrieve a real number answer from this? – akot717 Nov 14 '17 at 2:23
• Are you taking the absolute value inside of the logarithm? You should not have a complex valued result. – aleden Nov 14 '17 at 2:29
• Just edited, one of the values for the $\arcsin$ had a domain error, which would give you complex valued results. However, it is evident from the graph that the roots are just negatives of each other. – aleden Nov 14 '17 at 2:40

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

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