0
$\begingroup$

Let $V$ be the region in $\mathbb{R}^3$ satisfying inequalities $x^2+y^2 \le 1, 0 \le z \le 1$. Sketch $V$ and calculate:

$$\displaystyle \int \int \int_{V}(x^2+y^2+z^2)\,dV$$

I wrote the integral as $\displaystyle \int \int \int_{V}(x^2+y^2+z^2)\,dV = 4 \int_0^1 \int_0^{\sqrt{1-y^2}}\int_{0}^{1}(x^2+y^2+z^2)\,{dz}\,{dx}\,{dy}$.

Now I want to use polar coordinates for $x$ and $y$, so I let $x = r\cos{\theta}, y= r\sin{\theta}$. But what's the upper bound $\sqrt{1-y^2}$ in polar coordinates? I can't reduce it to something not dependent on $r$.

$\endgroup$
2
  • $\begingroup$ Hint: Make $x=r\cos(\eta)$ and $y=r\sin(\eta)$, where $0\leq r\leq 1$ and $0\leq\eta\leq 2\pi$. $\endgroup$
    – Rafael
    Commented Apr 13, 2016 at 0:47
  • $\begingroup$ I think I get. Correct if I'm wrong but it's $0 \le r \le \max(\sqrt{1-r^2\sin^2(\theta) }) \iff 0 \le r \le 1 $. So $\sqrt{1-y^2}$ becomes $1$. $\endgroup$
    – user129566
    Commented Apr 13, 2016 at 1:08

2 Answers 2

Reset to default
2
$\begingroup$

Probably it is faster to avoid polar coordinates and just exploit the simmetries of the cylinder $V$:

$$\begin{eqnarray*}\iiint_V (x^2+y^2+z^2)\,d\mu &=& \iiint_V (x^2+y^2)\,d\mu + \iiint_V z^2\,d\mu\\&=&\iint_{x^2+y^2\leq 1}(x^2+y^2)\,dx\,dy+\pi\int_{0}^{1}z^2\,dz\\&=&\int_{0}^{1}\rho^2\cdot2\pi\rho\,d\rho+\frac{\pi}{3} \\&=&\color{red}{\frac{5\pi}{6}}.\end{eqnarray*}$$

$\endgroup$
1
$\begingroup$

Notice that this is the volume of a cylinder. I would use cylindrical coordinates. You can describe this region equivalently as $0\leq r\leq 1$, $0\leq z \leq1$, $0\leq\theta\leq2\pi$. Then you can take the triple integral as $$\int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1} r \ dzdrd\theta$$ and solve from there.

$\endgroup$
3
  • $\begingroup$ What you found is $\iiint_V dV$, and $x^2+y^2+z^2$ term is missing. $\endgroup$
    – choco_addicted
    Commented Apr 13, 2016 at 1:28
  • $\begingroup$ My question was more on why $\sqrt{1-y^2}$ becomes $1$ in polar coordinates because if you let $y = r\sin{\theta}$ it just becomes $\sqrt{1-r^2\sin^2{\theta}}$, not $1$. Though not with certainty, I think it's because we consider the maximum of this, which is $1$ and that becomes the upper bound for $r$. $\endgroup$
    – user129566
    Commented Apr 13, 2016 at 1:32
  • $\begingroup$ It is not the volume of the cylinder, the volume is $\iiint_V 1\,d\mu$, but here we have $\iiint_V (x^2+y^2+z^2)\,d\mu$, that is something related with a moment of inertia. $\endgroup$
    – Jack D'Aurizio
    Commented Apr 13, 2016 at 2:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .