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I am working on a exercise where I have to integrate $f(x,y) = x+y$ over a cardioid. Ofcourse, using polar coordinates we can get there. Now, I have to calculate the double integral in both ways, so: $$ \int_0^\pi \int_0^{1+\cos\theta} r^2(\cos\theta + \sin\theta)\text{d} r \text{d} \theta$$ And $$\int_0^2\int_0^{\arccos(r-1)}r(\cos\theta+\sin\theta)r\text{d}\theta\text{d}r$$ I already calulcated the first one, and the resulting volume of this integral is $\frac{5\pi}{4}$. Now I'm trying to do the second one, but after calculated the inner integral I get stuck. I also show the steps I made, just in case: \begin{align*} &\int_0^2\int_0^{\arccos(r-1)}r(\cos\theta+\sin\theta)r\text{d}\theta\text{d} r\\ &= \int_0^2 r^2 \int_0^{\arccos(r-1)} \cos\theta+\sin\theta\text{d}\theta\text{d}r\\ &= \int_0^2 r^2 \left[\sin\theta - \cos\theta\right]_0^{\arccos(r-1)}\text{d}r\\ &= \int_0^2 r^2 \left(\sin(\arccos(r-1) - \cos(\arccos(r-1)) -\sin(0) + \cos(0)\right)\text{d} r\\ &= \int_0^2 r^2 \left(\sin(\arccos(r-1)) -r +1 -0 +1\right)\text{d} r \end{align*} Using that $\sin(\arccos x) = \sqrt{1 - x^2}$ I simplified the expression to $$ \int_0^2 r^2 (\sqrt{1-(r-1)^2} - r + 2)\text{d} r$$

But I really can't get a proper answer from this. Mathematica suggests a quite complicated integral, which I find hard to believe that that is answer I should get. Is there a clever way to solve this integral or did I make a mistake on my way to this integral

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$$\int _0^2\int _{-\arccos (r-1)}^{\arccos (r-1)}r^2(\cos \theta +\sin \theta )d\theta dr=$$

$$\int _0^22r^2\sqrt{2r-r^2}dr=$$

use a substitution $r=1+cos u$ to get

$$\int _0^{\pi }2(1+\cos u)^2\sin ^2udu=$$

Expand and use a half angle formula to get the result $$\frac{5 \pi }{4}$$

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  • $\begingroup$ Why would this be trivial by any way? $\endgroup$ – Rich_Rich Feb 27 '17 at 20:26
  • $\begingroup$ @RichardDirven I edited my answer above $\endgroup$ – Lozenges Feb 27 '17 at 21:39
  • $\begingroup$ Ahh yes that's what I found out too, in the end. Nonetheless thanks! Mathematica came up with an answer full of logarithms.. $\endgroup$ – Rich_Rich Feb 28 '17 at 7:30

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