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Difficult double integration problem

fanttamdiv

New member
Oct 16, 2013
3
I am getting a different answer from my colleague on the double integral below so I am trying to find out why. I am trying to put this into excel so I can play with the numbers and see the different results. The double integral is below:
DoubleIntegral.png

So the first integration (with respect to y) yields me this:
​\[\int_{-c}^{c}\frac{(x^2*\sinh^{-1}{\frac{\sqrt{a^2-x^2}}{x}})-(b*\sqrt{x^2+b^2})+(a*\sqrt{a^2-x^2})-(x^2*\sinh^{-1}{\frac{b}{x}})}{2} dx\]
(Online calc input):
(.5)*(((x^2)*(arsinh((sqrt((a^2)-(x^2)))/x)))-(b*(sqrt((x^2)+(b^2))))+(a*(sqrt((a^2)-(x^2))))-((x^2)*(arsinh(b/x))))

Remember, I am putting this through excel so I am keeping the variables in there until the end. The second integration (with respect to x now) yields me this:
\[\frac{(-b^3*\sinh^{-1}{\frac{c}{b}})+(a^3*\sin^{-1}\frac{c}{a})-(bc*\sqrt{c^2+b^2})+(ac*\sqrt{a^2-c^2})}{2}\]
(Online calc input):
.5*(((-b^3)*(arsinh(c/b)))+((a^3)*(arcsin(c/a)))-((b*c)*(sqrt((c^2)+(b^2))))+((a*c)*(sqrt((a^2)-(c^2)))))

So now that I have this, I can use excel to plug in various variables and see how they affect my answer. The current variables I am using are:
a: 2
b: 1.2
c: 1

With the variables above, I am getting ~2.234. My colleague, however, is getting ~2.361. I do not have access to his work, and he seems confident in his answer (his answer makes more sense to me too based on some other details I haven't explained here). Basically I am bringing up this issue a couple years after he did the original problems and he isn't allotted any more time to work on this so I'm on my own. Is there something I am doing wrong? I don't think it is a rounding error as I have gotten the same answer with Excel and an online integration calculator. Also, I have tried to get this question answered on a different forum and it was suggested that I convert the integral to polar coordinates. It's been way too long since I took calculus and I don't know how to do this... Anyone think they can help do this/figure out why I'm getting a different answer?

Thanks
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I have a few comments:

1. Am I right in thinking your integral is
$$ \int_{-c}^{c} \int_{b}^{ \sqrt{a^{2}-x^{2}}} \sqrt{x^{2}+y^{2}} \, dy \, dx?$$

2. If so, then may I ask whether your region in the $xy$ plane satisfies the relationship, in general, that $c^{2}+b^{2}=a^{2}$? Or is $c^{2}+b^{2}<a^{2}$? Or is $c^{2}+b^{2}>a^{2}$? (I should think the last option highly unlikely, as you will get complex numbers floating around!)

3. If we go on the assumption that $c^{2}+b^{2}=a^{2}$, then the integral becomes, in polar coordinates,
\begin{align*}
\int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \int_{b \csc( \theta)}^{a}r^{2} \, dr \, d \theta &=
\int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \left[ \frac{r^{3}}{3} \right]_{b \csc( \theta)}^{a} \, d \theta
\\
&= \frac{1}{3} \int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \left[a^{3}-b^{3} \csc^{3}( \theta) \right] \, d \theta \\
&= \frac{1}{3} \Bigg[ a^{3} \theta-b^{3} \Bigg( - \frac{1}{4} \ln \left|1+ \cos( \theta) \right|+ \frac{1}{4} \ln \left|1- \cos( \theta) \right| \\
& \qquad - \frac{1}{4(1- \cos( \theta))}+ \frac{1}{4(1+ \cos( \theta))}\Bigg) \Bigg]_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \\
&= \frac{2}{3} \, a^{3} \arcsin(b/a).
\end{align*}
 

fanttamdiv

New member
Oct 16, 2013
3
I have a few comments:

1. Am I right in thinking your integral is
$$ \int_{-c}^{c} \int_{b}^{ \sqrt{a^{2}-x^{2}}} \sqrt{x^{2}+y^{2}} \, dy \, dx?$$

2. If so, then may I ask whether your region in the $xy$ plane satisfies the relationship, in general, that $c^{2}+b^{2}=a^{2}$? Or is $c^{2}+b^{2}<a^{2}$? Or is $c^{2}+b^{2}>a^{2}$? (I should think the last option highly unlikely, as you will get complex numbers floating around!)

3. If we go on the assumption that $c^{2}+b^{2}=a^{2}$, then the integral becomes, in polar coordinates,
\begin{align*}
\int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \int_{b \csc( \theta)}^{a}r^{2} \, dr \, d \theta &=
\int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \left[ \frac{r^{3}}{3} \right]_{b \csc( \theta)}^{a} \, d \theta
\\
&= \frac{1}{3} \int_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \left[a^{3}-b^{3} \csc^{3}( \theta) \right] \, d \theta \\
&= \frac{1}{3} \Bigg[ a^{3} \theta-b^{3} \Bigg( - \frac{1}{4} \ln \left|1+ \cos( \theta) \right|+ \frac{1}{4} \ln \left|1- \cos( \theta) \right| \\
& \qquad - \frac{1}{4(1- \cos( \theta))}+ \frac{1}{4(1+ \cos( \theta))}\Bigg) \Bigg]_{ \arcsin(-b/a)}^{ \arcsin(b/a)} \\
&= \frac{2}{3} \, a^{3} \arcsin(b/a).
\end{align*}
Thanks for taking the time to reply! Your first assumption is correct, that is the double integral I am working with. I am afraid variables a, b, and c are not related in any of the ways you mentioned, i'm not even sure if they have an absolute relationship. Just treat a, b, and c as independent variables. I have attached below a picture of what a, b, and c stand for so you can visualize the shape I am working with (ignore the other variables in yellow). This is part of a larger problem and I am fairly confident this double integral is the source of my error. EDIT: The picture in the attachment is wrong, the picture that is embedded is correct.

Brake Pad Area (Shaded2).png

Also, I don't NEED it to be in polar coordinates, I just need a way to get an integrated solution that has variables I can change and see how it affects the final solution. Thanks for your help!
 

Attachments

Last edited:

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I am getting a different answer from my colleague on the double integral below so I am trying to find out why. I am trying to put this into excel so I can play with the numbers and see the different results. The double integral is below:
View attachment 1496
...

I can use excel to plug in various variables and see how they affect my answer. The current variables I am using are:
a: 2
b: 1.2
c: 1

With the variables above, I am getting ~2.234. My colleague, however, is getting ~2.361.
As someone on another forum suggested, you can evaluate the integral \(\displaystyle \int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx\) by transforming it to polar coordinates. You have to split it into two pieces, as in this diagram:

template.png

Let $d = \sqrt{b^2+c^2}$, $e = \sqrt{a^2-c^2}$, $\alpha = \arccos(b/d)$ and $\beta = \arcsin(c/a)$, as indicated in the diagram. Then $$\int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \int_b^d\int_{-\arccos(b/r)}^{\arccos(b/r)}r^2\,d\theta dr + \int_d^a\int_{-\arcsin(c/r)}^{\arcsin(c/r)}r^2\,d\theta dr.$$ The theta integrals are easy, and lead to $$\int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \int_b^d 2r^2\arccos(b/r)\,dr + \int_d^a 2r^2\arcsin(c/r)\,dr.$$ Those integrals are doable but not so quick. I'll spare you the details, but briefly what you have to do is to integrate by parts, then use the substitution $r = b\sec\varphi$ for the first integral and $r = c\csc\psi$ for the second one. The result comes out as $$\boxed{\displaystyle \int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \tfrac23a^3\beta + \tfrac13ace - \tfrac23bcd + \tfrac13c^3\ln\Bigl(\frac{d-b}{a-e}\Bigr) - \tfrac13b^3\ln\Bigl(\frac{c+d}{b}\Bigr)}.$$

Putting in the values $a=2$, $b=1.2$, $c=1$ (and remembering that $\beta$ must be measured in radians!), that formula gives the value of the integral as 2.361. So I'm sorry to say that it looks as though your colleague is right and you are wrong.
 
Last edited:

fanttamdiv

New member
Oct 16, 2013
3
As someone on another forum suggested, you can evaluate the integral \(\displaystyle \int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx\) by transforming it to polar coordinates. You have to split it into two pieces, as in this diagram:


Let $d = \sqrt{b^2+c^2}$, $e = \sqrt{a^2-c^2}$, $\alpha = \arccos(b/d)$ and $\beta = \arcsin(c/a)$, as indicated in the diagram. Then $$\int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \int_b^d\int_{-\arccos(b/r)}^{\arccos(b/r)}r^2\,d\theta dr + \int_d^a\int_{-\arcsin(c/r)}^{\arcsin(c/r)}r^2\,d\theta dr.$$ The theta integrals are easy, and lead to $$\int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \int_b^d 2r^2\arccos(b/r)\,dr + \int_d^a 2r^2\arcsin(c/r)\,dr.$$ Those integrals are doable but not so quick. I'll spare you the details, but briefly what you have to do is to integrate by parts, then use the substitution $r = b\sec\varphi$ for the first integral and $r = c\csc\psi$ for the second one. The result comes out as $$\boxed{\displaystyle \int_{-c}^c\int_b^{\sqrt{a^2-x^2}}\sqrt{x^2+y^2}\,dydx = \tfrac23a^3\beta + \tfrac13ace - \tfrac23bcd + \tfrac13c^3\ln\Bigl(\frac{d-b}{a-e}\Bigr) - \tfrac13b^3\ln\Bigl(\frac{c+d}{b}\Bigr)}.$$

Putting in the values $a=2$, $b=1.2$, $c=1$ (and remembering that $\beta$ must be measured in radians!), that formula gives the value of the integral as 2.361. So I'm sorry to say that it looks as though your colleague is right and you are wrong.
Brilliant, exactly what I was looking for! Thank you very much!:D