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I am having a difficult time solving this problem. I have tried this several different ways, and I get a different result, none of which is correct, every time. I've derived an answer geometrically and cannot replicate it with a double integral.

Here's the problem: Use a double integral to find the area between two circles $$x^2+y^2=4$$ and $$(x−1)^2+y^2=4.$$

Here is how I have tried to go about this problem:

First, I graphed it to get a good idea visually of what I was doing. Here's the graph I scribbled on. The region I'm interested is where these two circles overlap. This region can easily be divided into two separate areas. There are clearly a number of ways to go about solving this...but the one I opted for is to find the shaded region. The bounds for $x$ in this case are between $D$ and $C$. D can be found by setting $C_1=C_2$, and $x$ turns out to be $\frac{1}{2}$. On the right, $x$ is where $C_1(y)=0$, $x=\pm2$, so $x=2$ at point $C$. $y$ is greater than $B_y$ and less than $A_y$, which are also found where $C_1=C_2$, and $y$ turns out to be $\pm\sqrt{\frac{15}{4}}$. So far so good. Now I know my limits of integration. But here's what I don't understand. What am I actually integrating? $x$ has constant bounds, and $y$ does not, and looking at other double integral problems, that would lead me to believe that I should integrate $y$ first as a function of $x$, evaluate it at its bounds, and then integrate $x$ and evaluate it at its bounds giving me half the area I am looking for. However, when I try to do this, I get utter nonsense for an answer, or I get lost trying to set up the problem.

I could really use the help, I've spent entirely too much time trying to puzzle through this. Thank you in advance!

P.s. I determined the area geometrically using a CAD program to calculate the area, and it should be approximately $8.46$.

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  • $\begingroup$ do you need by using the polar coordinates?? $\endgroup$ – E.H.E Oct 15 '15 at 7:31
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As a double integral, this is $I=\int_R dA$, where $R$ is the region given by the intersection of the two disks. This integral can be evaluated as an iterated integral in several ways. Picking up where you left off, we can integrate with respect to $y$ first, then with respect to $x$, i.e., $$I=2\int_\frac12^2 \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dy\;dx.$$ You could take further advantage of symmetry by changing the lower bound of $y$ to $0$ and multiplying by $2$, but that doesn’t really make things any simpler.

You could instead integrate with respect to $x$ first:$$I=4\int_0^{\frac{\sqrt{15}}2}\int_\frac12^{\sqrt{4-y^2}} dx\;dy.$$ The key thing in both cases is that the bounds of the inner integral will be variable. I often find it helpful to draw a typical line segment for the inner integral to make sure that I get the bounds right. For example, in this second integral, for each value of $y$, the integral runs over a line segment parallel to the $x$-axis that runs from $x=\frac12$ out to the edge of the left-hand disk.

If you don’t want to deal with iterated integrals at all, this problem is a good candidate for Green’s theorem: $I=\int_R dA=\int_{\partial R}y\;dx$. You can often make the line integral simpler without changing its value by adding the differential of some function to the integrand. If you parametrize the bounding arcs as $\langle 2\cos t,2\sin t\rangle$ and $\langle 1-2\cos t,-2\sin t\rangle$, respectively, and use the differential form $\frac12(x\;dy-y\;dx)$ for the line integral, the integrals end up being very simple to evaluate:$$I=\int_{-\arccos\frac14}^{\arccos\frac14}2\;dt+\int_{-\arccos\frac14}^{\arccos\frac14}2-\cos t\;dt=\int_{-\arccos\frac14}^{\arccos\frac14}4-\cos t\;dt.$$

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  • $\begingroup$ This gave me what I needed to solve the problem. I appreciate the different approaches you presented; I was able to verify my answer several different ways. Thanks! $\endgroup$ – Josiah Oct 15 '15 at 18:07
  • $\begingroup$ In this problem, one can easily skip setting up a double integral and go directly to the definite single integral $4\int_\frac12^2\sqrt{4-x^2}\;dx$ for the area, but that probably defeats the purpose of the exercise. $\endgroup$ – amd Oct 15 '15 at 18:59
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Here is a similar problem solved.

Here is an example of less similar problem, it is for volume between two balls and therefore, of course, it is using a triple integral.

Hope this helps.

ps: note the links

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by using the Cartesian coordinates $$\int_{-\frac{\sqrt{15}}{2}}^{\frac{\sqrt{15}}{2}}\int_{1-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dxdy$$

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Surface area $$=\int \int_S dS$$ $$=\int \int_S \frac{dx \,\ dy}{\hat n\cdot\hat k}$$ since the surface considered here is $z=0$ $$=\int^{y=2}_{y=-2} \int^{x=\sqrt{4-y^2}}_{x=1-\sqrt{4-y^2}} dx \,\ dy$$ $$=\int^{y=2}_{y=-2} (2\sqrt{4-y^2}-1)dy$$ $$= 4(\pi - 1)$$

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  • $\begingroup$ This result seems incorrect, since it doesn’t match the value of any of the integrals in the other answers (which do all agree with each other). $\endgroup$ – amd Oct 15 '15 at 9:53
  • $\begingroup$ @amd But it matches with the result that the OP has given. $\endgroup$ – SchrodingersCat Oct 15 '15 at 9:55
  • $\begingroup$ Does it? When I compute $4(\pi-1)$ using several different calculators and CAS programs, they all come up with $8.56637$, which is a fair way off from $8.46$. The value I get in Mathematica from directly computing the area of the region also jibes with the other integrals. $\endgroup$ – amd Oct 15 '15 at 10:03

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