Finding the volume between paraboloids, cylinder and planes

Question

I have to find the volume of the solid $\Omega$ which is bounded by the paraboloids $$z=x^{2}+y^{2},\quad z=4x^{2}+4y^{2}$$ by the planes $$y=\sqrt{3}x,\quad y=\frac{\sqrt{3}}{3}x$$ and by the cylinders $$x^{2}+y^{2}=1,\quad x^{2}+y^{2}=4$$ I draw it here: Omega and I really have no idea how to find the limits of integration for $x,y$ and $z$. Actually, I can't even see what region is $\Omega$.

Answer 1

What you have is: $$\cases{x^2+y^2\leq z\leq 4(x^2+y^2)\\\\\frac1{\sqrt{3}}x\leq y\leq x\sqrt 3 & if $x\geq 0$\\\\ x\sqrt 3\leq y\leq\frac1{\sqrt{3}}x & if $x\leq 0$\\\\1\leq x^2+y^2\leq 4}$$

So the bounds on $z$ depend on $x$ and $y$, we will deal with it last. The bound on $y$ naturally depends on $x$, so we deal with $x$ first.

The last inequality implies that $[-2,-1]\cup [1,2]$ is the projection of your domain on the $x$-axis. Say we focus on $x\geq 0$.

For a given $x$, $y$ is bound by the second inequality, but also the third (e.g. if $x=2$, $y$ has to be $0$): $$y\in \left[\frac1{\sqrt{3}}x, x\sqrt 3 \right]\cap \left[\sqrt{1-x^2},\sqrt{4-x^2}\right]$$ So you need to figure out the intersection between these intervals. It involves answering questions such as 'which of $\frac 1{\sqrt{3}}x$ and $\sqrt{1-x^2}$ is greater'. The intersection is an interval $\left[y_{\min}(x), y_{\max}(x)\right]$.

In the end, you have to compute the integral $$\int_{x=1}^2 \int_{y=y_{\min}(x)}^{y_{\max}(x)}\int_{z={x^2+y^2}}^{4(x^2+y^2)}dz dy dx$$

Can you fill in the details?

Answer 2

Even if it is hard to draw, let us rewrite the conditions in a compact way: $$x^2+y^2\le z\le 4(x^2+y^2)\ ,$$ $$1\le x^2+y^2\le 4\ ,$$ $$\frac 1{\sqrt 3}\le \frac xy\le \sqrt 3\ .$$ So the volume of $\Omega$, the solid delimited by the above inequalities, can be computed via Fubini as follows: $$ \begin{aligned} \operatorname{Vol}(\Omega) &= \iint_{\substack{1\le x^2+y^2\le 4\\\sqrt{1/3}\le x/y\le \sqrt 3}} dx\;dy \int_{x^2+y^2}^{4(x^2+y^2)}\; dz \\ &= \iint_{\substack{1\le x^2+y^2\le 4\\\sqrt{1/3}\le x/y\le \sqrt 3}} \Big(\ 4(x^2+y^2)-(x^2+y^2)\ \Big)\;dx\;dy \\ &= 3\iint_{\substack{1\le x^2+y^2\le 4\\\sqrt{1/3}\le x/y\le \sqrt 3}} (x^2+y^2)\;dx\;dy \\ &= 6\iint_{\substack{1\le x^2+y^2\le 4\\\sqrt{1/3}\le x/y\le \sqrt 3\\x,y>0}} (x^2+y^2)\;dx\;dy \\ &\qquad\text{now we pass to polar coordinates...} \\ &= 6\iint_{\substack{1\le r^2\le 4\\\pi/6\le t\le \pi/ 3\\r>0}} r^2\;r\;dr\;dt \\ &= 6\int_1^2 r^3\; dr \int_{\pi/6}^{\pi/ 3}dt \\ &= 6\cdot\frac 14(2^4-1^4)\cdot\left(\frac \pi3-\frac \pi6\right)\ . \\ &=\frac{15\pi}4 \end{aligned} $$

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Computer simulation, sage, using the fact that $\Omega$ is inside the box $[-2,2]\times[-2,2]\times[1,16]$ of known volume $V$:

V = 4*4*15
import random
N = 10**6    # trials
ok = 0    # so far
for trial  in xrange(N):
    x,y,z = random.uniform(-2,2), random.uniform(-2,2), random.uniform(1,16)
    rr, f = x^2+y^2, x/y
    if ( 1 <= rr and rr <= 4 and rr <= z and z <= 4*rr
         and f > 0 and f^2 <=3 and f^2 >= 1/3 ):
        ok += 1

print "Monte-Carlo volume this time ~ %s" % ( 1.0 * V * ok/N)
print "Computed volume = %s ~ %s" % ( 15*pi/4, (15*pi/4).n() )

Results:

Monte-Carlo volume this time ~ 11.7907200000000
Computed volume = 15/4*pi ~ 11.7809724509617