Complex double integration problem

[fanttamdiv]

Basically, 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:

The values I am using are as follows:

a: 2
b: 1.2
c: 1

So the first integration (with respect to y) yields me this:
​\[\int_{-c}^{c}\frac{(x^2*\sinh^{-1}{\frac{\sqrt{a^2-x^2}}{x}})-(b*\sqrt{x^2+b^2})+(a*\sqrt{a^2-x^2})-(x^2*\sinh^{-1}{\frac{b}{x}})}{2} dx\]
(Excel 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:
\[\frac{(-b^3*\sinh^{-1}{\frac{c}{b}})+(a^3*\sin^{-1}\frac{c}{a})-(bc*\sqrt{c^2+b^2})+(ac*\sqrt{a^2-c^2})}{2}\]
(Excel 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 have excel plug in the variables listed in the above section and get the answer. 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.

Thanks

 

[Deleted member 4993]

Basically, 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 3318
The values I am using are as follows:

a: 2
b: 1.2
c: 1

So the first integration (with respect to y) yields me this:
​\[\int_{-c}^{c}\frac{(x^2*\sinh^{-1}{\frac{\sqrt{a^2-x^2}}{x}})-(b*\sqrt{x^2+b^2})+(a*\sqrt{a^2-x^2})-(x^2*\sinh^{-1}{\frac{b}{x}})}{2} dx\]
(Excel 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:
\[\frac{(-b^3*\sinh^{-1}{\frac{c}{b}})+(a^3*\sin^{-1}\frac{c}{a})-(bc*\sqrt{c^2+b^2})+(ac*\sqrt{a^2-c^2})}{2}\]
(Excel 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 have excel plug in the variables listed in the above section and get the answer. 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.

Thanks

The best way to do this would be to convert to polar co-ordinates.

x = r*cos(Θ)

dy = r*sin(Θ)

dA = dy dx = r dr dΘ

Changing the limits can be a bit more difficult but can be done knowing that the region is part of a circle x²+ y² = a²

 

[fanttamdiv]

Thanks for your response. Wouldn't this net me the same answer though? Regardless, I'm pretty rusty on my calculus and was only able to complete these calculations with an online calculator. Any way you could give me a more in depth walk-through on converting these to polar coordinates?

Thanks