calculusdefinite integralsintegrationmultivariable-calculus
I need to calculate this volume of
$$D=\{x^2+y^2-2y\le 0,0\le z\le 10-3\sqrt{x^2+y^2} \}.$$
My attempt. So the first one is a shifted cylinder $x^2+(y-1)^2\le 1$ , and the second one is an upside-down cone with vertex $z=10$.
By using cylindrical coordinates I get:
$$\int_{0}^{2\pi}\int_{0}^{\sqrt{2\sin{\theta}}}\int_{0}^{10-3r}rdzdrd\theta$$
Is this the correct way to approach a shifted region? $x^2+y^2=2y$ implies $r^2=2\sin{\theta}$.
I think I got the integral wrong.
Answer using Cylindrical Coordinates:
Volume of the Shared region =
Equating both the equations for z, you get z = 1/2. Now substitute z = 1/2 in in one of the equations and you get r = $\sqrt{\frac{3}{4}}$.
Now the sphere is shifted by 1 in the z-direction, Hence
Volume of the Shared region = $$\int_{0}^{2\pi} \int_{0}^{\sqrt{\frac{3}{4}}} \int_{1-\sqrt{1-r^2}}^{\sqrt{1-r^2}} rdzdrd\theta$$
$$V=2\pi \int_{0}^{\sqrt{\frac{3}{4}}} [2{\sqrt{1-r^2}}-1] rdr$$
substitute $$u = 1-r^2 ; r = 0 => u = \frac{1}{4} ; r = \sqrt{\frac{3}{4}} => u = 1$$
$$V = 2\pi [-\int_1^{\frac{1}{4}} u^{\frac{1}{2}} du - \int_{0}^{\sqrt{\frac{3}{4}}} rdr]$$
$$V= 2\pi (\frac{2}{3}u^{\frac{3}{2}}) - (\frac{r^2}{2})$$ $$V =2\pi*( \frac{2}{3}(1-\frac{1}{8}) - \frac{3}{8})$$
$$V = 2\pi*(\frac{14}{24} - \frac{3}{8}) = 2\pi*\frac{5}{24} = \frac{5}{12} \pi$$
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 $J_x(y)=\left [\ -\sqrt{9-y^2},\ +\sqrt{9-y^2}\ \right]$.
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 $$ \begin{aligned} &\iiint_D\sqrt{10+y^2-x^2}\;dx\;dy\;dz \\ &\qquad= \int_{(x,y)\in\text{Disk of radius $3$ centered in origin}}dx\;dy \int_{z\in J_z(x,y)}\sqrt{10+y^2-x^2}\;dz \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \int_{z\in [-\sqrt{10-r^2\cos 2t},\ +\sqrt{10-r^2\cos 2t}]} \sqrt{10-r^2\cos 2t}\;dz \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot(\text{ length of $[-\sqrt{10-r^2\cos 2t},\ +\sqrt{10-r^2\cos 2t}]$ })\cdot \sqrt{10-r^2\cos 2t} \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot2(10-r^2\cos 2t) \\ &\qquad= \int_{(r,t)\in[-3,3]\times[0,2\pi]}r\;dr\;dt \cdot2\cdot 10 \\ &\qquad\qquad\text{ since the integral of $\cos 2t$ vanishes on $[0,2\pi]$,} \\ &\qquad= \int_0^32\pi\cdot r\;dr \cdot2\cdot 10 \\ &\qquad= 2\pi\cdot \frac 92 \cdot2\cdot 10 \\ &\qquad= 180\pi\ . \end{aligned} $$
Best Answer
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?