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I am trying to calculate the triple integral shown below. I've also computed the rectangular limits for x, y and z. I've attempted to compute the integral while staying in rectangular, and it ended up being several pages of wild integrals that just kept getting larger and larger. I tried to set this up to do as a cylinder, but I am getting lost. I feel like this problem should be easier, but I just can't wrap my head around it:

Problem statement and my calculation of the limits here.

In the above image, I've set up the integral correctly for rectangular (I think?) but actually computing it has proven to be overly labor intensive. Can someone please help me set up this integral in a better way? Thank you in advance!

P.s. I apologize for not using MathJax; I wanted to be able to write some nicely formatted math, but the learning curve looks a little high for me right now.

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METHOD 1: Cartesian Coordinates

We can exploit symmetry to evaluate this integral. First, we note that we can write the integral of interest $I$ as

$$I=\int_{-2}^2\int_{-2}^2 \int_{-\sqrt{4-y^2}}^\sqrt{4-y^2} \left(4+5x^2yz^2\right)\,dx\,dy\,dz$$

We proceed to evaluate the integral as

$$\begin{align} I&=\int_{-2}^2\int_{-2}^2 \left(8\sqrt{4-y^2}+\frac{10}3 (4-y^2)^{3/2}yz^2\right)\,dy\,dz \tag 1\\\\ &=64\int_0^2\sqrt{4-y^2}\,dy \\\\ &+\frac{160}{9}\int_0^2y(4-y^2)^{3/2}\,dy \tag 2\\\\ &=64\int_0^2\sqrt{4-y^2}\,dy \tag 3\\\\ &=64\left.\left(\frac{y\sqrt{4-y^2}}{2}+2\arctan\left(\frac{y}{\sqrt{4-y^2}}\right)\right)\right|_{0}^{2} \tag 4\\\\ &=64\pi \tag 4 \end{align}$$

In arriving at $(1)$, we carried out the integration over the inner integral (i.e., $x$).

In going from $(1)$ to $(2)$, we carried out the integral over $z$.

In going from $(2)$ to $(3)$ we exploited the fact that the integrand of the second integral in $(2)$ was an odd function of $y$ and the limits of integration were symmetric around $y=0$. Therefore, this integral is zero.

In arriving at $(4)$, we used standard trigonometric substitution $y\to 2\sin \theta$ to derive the anti-derivative of $\sqrt{4-y^2}$.

And finally, we used the fact that $\lim_{x\to 0^+}\arctan (1/x)=\pi/2$ to obtain $(5)$.


METHOD 2: Cylindrical Coordinates

We can express $I$ in terms of cylindrical coordinates as

$$I=\int_{-2}^2\int_0^{2\pi}\int_0^2\left(4+5\rho^3\cos^2\phi \sin \phi z^2\right)\,\rho \,d\rho\,d\phi\,dz$$

We see immediately that the integral of the second term of the integrand vanishes since $\int_0^{2\pi}\cos^2\phi \sin \phi \,d\phi=0$. Therefore, we have

$$I=\int_{-2}^2\int_0^{2\pi}\int_0^2\,4\,\rho \,d\rho\,d\phi\,dz=64\pi$$

as expected!

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  • $\begingroup$ Your answer is excellent. I am working my way through the problem now, and have run into a couple of stumbling points: 1) When integrating Z in the Cartesian method, the second term does not cancel out for me as it becomes positive and you get 2*<second term>. 2) The polar form looks much simpler to solve, but I am unsure how you converted from Cartesian to polar. $\endgroup$ – Josiah Oct 16 '15 at 17:19
  • $\begingroup$ Good Questions. First, you are right; well done! For the integration over $z$, the integral $\int_{-2}^2 5x^2yz^2\,dz=5x^2y\frac{16}{3}\ne 0$. It is only when we integrate over $x$ and $y$ that this term vanishes. This is due to the fact that $x^2y$ is an odd function of $y$ and we are integrating over symmetric limits. For the cylindrical coordinate transformation, we have $x=\rho \cos \phi$, $y=\rho \sin \phi$, $dx\,dy\to \rho\,d\rho\,d\phi$, and the new limits are $\phi \in[0,2\pi]$ and $\rho \in [0,2]$. $\endgroup$ – Mark Viola Oct 16 '15 at 17:26

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