0
$\begingroup$

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.

I didn't shade it, but the area they are asking for is the area above the red line (y=8) and enclosed by the polar curve

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.

$\endgroup$
  • 1
    $\begingroup$ 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?) $\endgroup$ – Ted Shifrin Nov 14 '17 at 1:51
  • $\begingroup$ 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 $16+16\sin\theta$? $\endgroup$ – akot717 Nov 14 '17 at 2:00
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I should probably know the closed forms for the $\arcsin$ expressions but I'm kinda blanking on them right now. $\endgroup$ – aleden Nov 14 '17 at 2:16
  • $\begingroup$ Thank you!! Makes sense now. I didn't think to use the quadratic formula for some reason. $\endgroup$ – akot717 Nov 14 '17 at 2:16
  • $\begingroup$ 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? $\endgroup$ – akot717 Nov 14 '17 at 2:23
  • $\begingroup$ Are you taking the absolute value inside of the logarithm? You should not have a complex valued result. $\endgroup$ – aleden Nov 14 '17 at 2:29
  • 1
    $\begingroup$ 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. $\endgroup$ – aleden Nov 14 '17 at 2:40
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.