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I'm trying to evaluate the following integral:

$\int\int\int_A z\ dxdydz$, where $A:=\{(x,y,z)\in\mathbb{R}^3:0\leq z\leq 1-\sqrt{x^2+y^2}, z\geq 1-4\sqrt{x^2+y^2}\}$ using cylindrical coordinates; I get $\int_{0}^{2\pi}\int_{1/4}^{1}\int_{1-4\rho}^{1-\rho}z\rho\ dzd\rho d\theta$ but when I evaluate it I get a negative result. What am I getting wrong in setting up this integral?

Best regards,

lorenzo.

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    @lorenzo Do you have the final result? I think this must be a split integral as we have two cones here, one into the other one but with the same vertex at $\;(0,0,1)\;$ – 2017-01-16
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    @b00n heT : why is that? – 2017-01-16
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    @b00nheT I don't think that's right. Read my past comment. – 2017-01-16
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    @DonAntonio: I thought that too, so I tried with the following: $I=I_1-I_2$, where $I_1=\int_{0}^{2\pi}\int_{0}^{1}\int_{0}^{1-\rho}z\rho\ dzd\rho d\theta$ and $I_2=\int_{0}^{2\pi}\int_{0}^{1/4}\int_{0}^{1-4\rho}z\rho\ dzd\rho d\theta$ and I got a positive result ($\approx 0.245$) but I don't have the final result so I don't know if that's right. – 2017-01-16
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    @lorenzo Read my answer. I think it is right. – 2017-01-16

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I think this integral has to be done in two parts: for the first one we get

$$0\le\theta\le2\pi\;,\;\;\frac14\le\rho\le1\;,\;\;0\le z\le1-\rho$$

and for the second part we have

$$0\le\theta\le2\pi\;,\;\;0\le\rho\le\frac14\;,\;\;1-4\rho\le z\le1-\rho$$

and we get:

$$\int_0^{2\pi}\int_{1/4}^1\int_0^{1-\rho}z\rho \,dz\, d\rho\, d\theta=\pi\int_{1/4}^1\rho(1-\rho)^2\,d\rho=\pi\int_{1/4}^1\left(\rho-2\rho^2+\rho^3\right)\,d\rho=$$

$$=\left.\pi\left(\frac12\rho^2-\frac23\rho^3+\frac14\rho^4\right)\right|_{1/4}^1=\pi\left(\frac12\left(1-\frac1{16}\right)-\frac23\left(1-\frac1{64}\right)+\frac14\left(1-\frac1{256}\right)\right)=$$

$$=\pi\left(\frac{15}{32}-\frac{21}{32}+\frac{255}{1,024}\right)=\frac{-192+255}{1,024}\pi=\frac{63}{1,024}\pi$$

and for the second integral:

$$\int_0^{2\pi}\int_0^{1/4}\int_{1-4\rho}^{1-\rho} \rho\,z\,dz\,d\rho\,d\theta=\pi\int_0^{1/4}\left(\rho(1-\rho)^2-\rho(1-4\rho)^2\right)\,d\rho=$$

$$=\pi\int_0^{1/4}\left(6\rho^2-15\rho^3\right)\,d\rho=\left.\pi\left(2\rho^3-\frac{15}4\rho^4\right)\right|_0^{1/4}=\pi\left(\frac1{32}-\frac{15}{1,024}\right)=\frac{17}{1,024}\pi$$

Adding both integrals, the final result I get is: $\cfrac5{64}\pi\;$

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    shouldn't you subtract the second integral from the first? and shouldn't the starting point of the $z$ part be $\int_{0}^{1-4\rho}$ in the second integral? – 2017-01-16
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    @lorenzo Perhaps you are right, but it depends on the interpretation of the given domain. I interpreted as meaning : all the domain above the plane $\;z=0\;$ and below the cone $\;1-\sqrt{x^2+y^2}\;$ (and this *already* contains the zone above the annulus $\;\frac14\le r\le 1\;$ !), and also below this conus and the conus $\;1-4\sqrt{x^2+y^2}\;$ . But now that you mention it, it could perfectly well be that they meant **only** the zone between both cones *and* above the plane $\;z=0\;$ , so then we *only* need the second integral (no need to substract) – 2017-01-16