Set up this triple integral

calculusdefinite integralsintegrationmultivariable-calculus

$D=${$ x^2-4(y^2+z^2)\ge{0} \ ,2x^2+y^2+z^2 \le{1} $}

So I have a double cone on the x axes
and an Ellipsoid

I have to compute $\int_{D}|x|(1+yz)dxdydz$

[UPDATE]

I think this is the way to proceed:

$\int_{D}|x|(1+yz)dV = \int_{D}x(1+yz)dV-\int_{D}x(1+yz)dV$

for the first integral, I took the x range :

$x=(range)= [2\sqrt{y^2+z^2},\sqrt{1-y^2/2-z^2/2}]$

In order to simplify the integral I change to cylindrical coordinates:

$x_r =(range)=[2r,\sqrt{1-r^2}/\sqrt{2}]$

eventually : ($1/3$ is the radius of the circle about the intersection of the cone and the ellipsoid)

$\int_{0}^{2\pi}\int_{0}^{1/3}\int_{2r}^{\sqrt{1-r^2}/\sqrt{2}}{x(1+r^2\sin{\theta}\cos{\theta})rdxdrd\theta}$

the integral seems kinda wrong, what do you think ?

Best Answer

Notice that by symmetry of $D$, we have

$$\int_D |x|yz \, dV = 0$$

as the signs of $y$ and $z$ will cancel each other out over different octants.

Hence the quantity of interest is just \begin{align}\int_D |x| \, dV &\end{align}

We can focus on just on the first octant and then multiply it by $8$.

\begin{align} &\int_D |x| \, dV \\&= 8 \int_{0}^{\frac{\pi}2}\int_{0}^{\frac13}\int_{2r}^{\sqrt{\frac{1-r^2}2}}xr\,\,\, dx dr d\theta \\ &=4 \int_{0}^{\frac{\pi}2}\int_0^\frac13r\left(\frac{1-r^2}{2} -4r^2\right)\,\, dr d\theta \\ &=4 \int_{0}^{\frac{\pi}2}\int_0^\frac13r\left(\frac{1-9r^2}{2} \right)\,\, dr d\theta \\ &=\pi \int_0^\frac13r\left(1-9r^2 \right)\, dr \\ &= \pi \left[ \frac{r^2}2-\frac{9r^4}{4}\right]_0^\frac13\\ &= \pi \left[ \frac1{18}-\frac1{36} \right]\\ &= \frac{\pi}{36} \end{align}

Related Solutions

Calculate this triple integral over D

First of all we see from $x^2+y^2=9$ that $y$ lives in $J_y=[-3,3]$.

Fix $y$ in $J_y=[-3,3]$. Then $x$ lives in the interval $J_x(y)=\left [\ -\sqrt{9-y^2},\ +\sqrt{9-y^2}\ \right]$.
- Fix now also $x$ in the above allowed interval $J_x(y)$. Then $y^2+10-x^2\ge y^2-10-9=y^2-1\ge 1>0$, so we always have a section of $D$ in the line with the fixed values $x,y$ for the first two coordinates, ant $z$ takes values in the interval $J_z(x,y)=\left[\ -\sqrt{y^2+10-x^2},\ +\sqrt{y^2+10-x^2}\ \right]$.

Now we can compute using Fubini: $$ \begin{aligned} &\iiint_D\sqrt{10+y^2-x^2}\;dx\;dy\;dz \\ &\qquad= \int_{y\in[-3,3]}dy \int_{x\in J_x(y)}dx \int_{z\in J_z(x,y)}\sqrt{10+y^2-x^2}\;dz \\ &\qquad= \int_{y\in[-3,3]}dy \int_{x\in J_x(y)}dx \cdot \left(\text{length of $J_z(x,y)$}\right)\cdot\sqrt{10+y^2-x^2} \\ &\qquad= \int_{y\in[-3,3]}dy \int_{x\in J_x(y)}dx \cdot 2\sqrt{10+y^2-x^2}\cdot\sqrt{10+y^2-x^2} \\ &\qquad= 2\int_{y\in[-3,3]}dy \int_{x\in J_x(y)}dx \cdot (10+y^2-x^2) \\ &\qquad= 2\int_{y\in[-3,3]}dy \cdot \left[\ 10x + xy^2-\frac 13x^3\ \right] _{x=-\sqrt{9-y^2}} ^{x=+\sqrt{9-y^2}} \\ &\qquad= 4 \int_{y\in[-3,3]}dy \cdot \sqrt{9-y^2}\cdot \left( \ 10 + y^2-\frac 13(9-y^2)\ \right) \\ &\qquad= 4 \int_{-3}^3 \sqrt{9-y^2}\cdot \left( \ 7 + \frac 43y^2\ \right)\; dy \\ &\qquad\qquad\text{Substitution: } y=3\sin t\ , \\ &\qquad= 4 \int_{-\pi/2}^{\pi/2} 3\cos t\cdot \left( \ 7 + \frac 43\cdot 9\sin^2 t\ \right)\; 3\cos t\; dt \\ &\qquad= 4\cdot 3^2\cdot 2 \int_0^{\pi/2} \cos^2 t\cdot \left( \ 7 + 12\sin^2 t\ \right)\; dt \\ &\qquad=180\pi\ . \end{aligned} $$

Computer check, here sage:

sage: integral( 
....:     sqrt(10+y^2-x^2), z, -sqrt(10+y^2-x^2), +sqrt(10+y^2-x^2) )
-2*x^2 + 2*y^2 + 20

sage: integral( integral( 
....:     sqrt(10+y^2-x^2), z, -sqrt(10+y^2-x^2), +sqrt(10+y^2-x^2) ),
....:   x, -sqrt(9-y^2), +sqrt(9-y^2) )
4/3*(4*y^2 + 21)*sqrt(-y^2 + 9)

sage: integral( integral( integral( 
....:     sqrt(10+y^2-x^2), z, -sqrt(10+y^2-x^2), +sqrt(10+y^2-x^2) ),
....:   x, -sqrt(9-y^2), +sqrt(9-y^2) ),
....: y, -3, 3 )
180*pi

Later edit:

Using cylindrical coordinates is possible, and this leads to a similar computation. I was avoiding this first because using $x=r\cos t$, $y=r\sin t$ we do not have an immediat simplification for $\sqrt{10+y^2-x^2}\ge 1$.

Note however that the correct condition for $z$, after we fix $x,y$ and/or after we fix the corresponding values for $r,t$, is $$ \begin{aligned} |z| &\le \sqrt{10+y^2-x^2}\ ,\text{ respectively}\\ |z| &\le \sqrt{10+r^2(\sin^2 t-\cos^2 t)} =\sqrt{10-r^2\cos2t} \ , \end{aligned} $$ so we would have to write $$ \begin{aligned} &\iiint_D\sqrt{10+y^2-x^2}\;dx\;dy\;dz \\ &\qquad= \int_{(x,y)\in\text{Disk of radius $3$ centered in origin}}dx\;dy \int_{z\in J_z(x,y)}\sqrt{10+y^2-x^2}\;dz \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \int_{z\in [-\sqrt{10-r^2\cos 2t},\ +\sqrt{10-r^2\cos 2t}]} \sqrt{10-r^2\cos 2t}\;dz \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot(\text{ length of $[-\sqrt{10-r^2\cos 2t},\ +\sqrt{10-r^2\cos 2t}]$ })\cdot \sqrt{10-r^2\cos 2t} \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot2(10-r^2\cos 2t) \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot2\cdot 10 \\ &\qquad\qquad\text{ since the integral of $\cos 2t$ vanishes on $[0,2\pi]$,} \\ &\qquad= \int_0^32\pi\cdot r\;dr \cdot2\cdot 10 \\ &\qquad= 2\pi\cdot \frac 92 \cdot2\cdot 10 \\ &\qquad= 180\pi\ . \end{aligned} $$

Evaluate this triple integral (volume)

You are almost correct. The angle $\theta$ should be in the interval $[0,\pi]$, so $0\leq r\leq 2\sin\theta\geq 0$ (note that $r^2\leq 2r\sin\theta$). Thus your triple integral should be $$\int_{0}^{\pi}\int_{0}^{2\sin{\theta}}\int_{0}^{10-3r}rdzdrd\theta.$$ So what is the final result?

Best Answer

Related Solutions

Calculate this triple integral over D

Evaluate this triple integral (volume)

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