Find the triple integral

$\iiint_{\mathrm{T}}x^{4}\, dxdydz$

where $T$ is bounded by $x^{2} + y^{2}+2y = 0$ and $z = 2x, z = 0$ planes.

My attempt at the solution:

$0\le z\le 2x, -2\le y\le 0, 0\le x\le \sqrt{-2y - y^{2}}$

$\int_{-2}^0dy\int_0^{\sqrt{-2y - y^{2}}}dx\int_0^{2x} x^{4}dz$

Is the solution correct? If not, then what is the correct solution?

  • $\begingroup$ I think $z$ can take both $x$ and $z$ can take negative values right? $\endgroup$ – Siong Thye Goh Apr 24 '16 at 23:27

There are two possible regions of equal volume bounded by $z=0$, $z=2x$, and $x^2+y^2+2y=0$, however, since $x^4$ is even in $x$, it doesn't matter which region, so I'll use the one with $x\geq 0$.

$$\iint_D\int_0^{2x}x^4dz dA=\iint_D2x^5dA$$

Now the region $D$ is a part of a circle that isn't centered at the origin, so we have to choose our limits of integration carefully. Converting $x^2+y^2+2y=0$ to polar coordinates gives $r=-2\sin\theta$. Since $z\in[0,2x]$, our region is the right side of the circle in the fourth quadrant. Integrating w.r.t. $\theta$ from $-\pi/2$ to 0 sweeps through that region and letting $r$ go from 0 to $-2\sin\theta$ completes the setup. This should give us a positive volume element. Now to evaluate the integral:

$$ \begin{aligned} \int_{-\pi/2}^0&\int_0^{-2\sin\theta}2r^6\cos^5\theta drd\theta \\ &= -\frac{2^8}{7}\int_{-\pi/2}^0 \sin^7(\theta)\cos^5(\theta)d\theta \\ &= -\frac{2^8}{7}\int_{-1}^0 t^7(1-t^2)^2dt \\ &= -\frac{2^8}{7}\int_{-1}^0 (t^7-2t^9+t^{11})dt \\ &= -\frac{2^8}{7}\left(\frac{1}{8}-\frac{1}{5}+\frac{1}{12}\right) \\ &=\frac{2^5}{7\cdot15}. \end{aligned} $$

  • $\begingroup$ Thank you for your answer. As I understand, my solution should be correct as well, as it gives the same answer if the integral is evaluated. $\endgroup$ – user02942473 Apr 25 '16 at 9:22
  • $\begingroup$ No problem. I like symbolab for checking work: symbolab.com/solver/triple-integrals-calculator $\endgroup$ – jdods Apr 25 '16 at 10:50

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