# Calculate this triple integral over D

$$\iiint \sqrt{10+y^2-x^2} \,dx \,dy \, dz$$ over the region bounded by $D$:

$D=\{(x,y,z)\mid x^2+z^2\le y^2+10,x^2+y^2\le 9\}$

So It's a cylinder intersecting a hyperboloid.

Is It correct to say that :

$$-3\le x \le 3$$

$$-3\le y \le 3$$

$$-\sqrt{2y^2-1} \le z \le \sqrt{2y^2-1}$$

?

And in the case, I want to use Cylindrical coordinates :

$$0\le \theta \le 2\pi$$

$$0 \le r \le 3$$

$$-\sqrt{2r^2\cos^2(\theta)-1}\le z \le \sqrt{2r^2cos^2(\theta)-1}$$

?

I think I got it wrong because the integral its kinda hard.

• And in the case I can use cylindrical coordinates , how can I simplify the $f(x,y,z)$ ?
• In Cartesian coordinates, your $x$ and $y$ seem fine, but the third equation should be $z^2\le 10+y^2-x^2$. As $y^2-x^2\ge-9$, this is the same as $-\sqrt{10+y^2-x^2}\le z\le \sqrt{10+y^2-x^2}$. – Kusma Jul 27 '18 at 17:08
• Yeah thanks you are right – NPLS Jul 27 '18 at 17:11

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}

• What would the cylindrical coordinates be if I wanted to use them? – NPLS Jul 27 '18 at 17:30
• Thank you so much! – NPLS Jul 28 '18 at 9:48
• How do you use sage to integrate in cylindrical coordinates? I tried integral(integral(integral( ....: 1,z,-3,3), ....: t,0,2*pi), ....: r,3,sqrt(18)) but 't' (theta) is not defined – NPLS Jul 28 '18 at 11:53