1
$\begingroup$

Determine the volume of the solid described by $$x^2 + y^2 + 2z^2 \leq 1; \, x + 2y + 3z \geq 0.$$ I am pretty sure that I will need to be doing a triple integral here, but I'm not quite sure how to set up the integrals. I am mostly unsure about the bounds of integration of each variable. I have also tried converting to spherical coordinates but it didn't seem to be helpful.

I was able to discern that the region in question would be an ellipsoid intersected by a plane and was able to verify my idea with Mathematica.

This is the region in question.

$\endgroup$
4
  • $\begingroup$ Which equation is ellipsoid here? $\endgroup$
    – Math Lover
    Jun 8, 2021 at 22:17
  • $\begingroup$ Pls edit your question to show your effort. Add more context. Why are you stuck? If the plane was $z \geq 0$ instead, would you know how to find volume of intersection? Do you know spherical coordinates? Do you know rotation of coordinate axes? $\endgroup$
    – Math Lover
    Jun 8, 2021 at 22:25
  • $\begingroup$ Do you have to use calculus? As the plane passes through the center of the sphere, the answer is obvious but is it a triple integral exercise? These are important contexts that your question is missing. $\endgroup$
    – Math Lover
    Jun 8, 2021 at 22:29
  • $\begingroup$ No calculus is not necessarily required, although I would like to use it for the sake of practice. Yes, as I mentioned I tried converting the equation to spherical but it did not seems to help. I actually typed the problem incorrectly. The first equation should end in $+2z^2$. $\endgroup$ Jun 8, 2021 at 23:54

3 Answers 3

2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson\ Bracket}$. \begin{align} V & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}} \bracks{x^{2} + y^{2} + 2z^{2} < 1}\bracks{x + 2y + 3z > 0} \dd x\,\dd y\,\dd z} \\[5mm] & = {\root{2} \over 2}\iiint_{\mathbb{R}^{3}} \bracks{x^{2} + y^{2} + z^{2} < 1} \bracks{x + 2y + {3\root{2} \over 2}\,z > 0}\dd x\,\dd y\,\dd z \\[5mm] & = \left.{\root{2} \over 2}\iiint_{r\ <\ 1} \bracks{\vec{r}\cdot\vec{n} > 0}\dd^{3}\vec{r} \,\right\vert_{\ \vec{\,\large n}\ \equiv\ \pars{1,2,3\root{2}/2}} \\[5mm] & = {\root{2} \over 2}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \bracks{rn\cos\pars{\theta} > 0}r^{2}\sin\pars{\theta} \,\dd r\,\dd\theta\,\dd\phi \\[5mm] & = {\root{2} \over 2}\,2\pi\int_{0}^{\pi}\int_{0}^{1} \bracks{\cos\pars{\theta} > 0}r^{2}\sin\pars{\theta} \,\dd r\,\dd\theta \\[2mm] & = \root{2}\pi\int_{0}^{\pi} \bracks{0 < \theta < {\pi \over 2}}\sin\pars{\theta} \pars{1 \over 3}\,\dd\theta = {\root{2} \over 3}\pi\ \overbrace{\int_{0}^{\pi/2}\sin\pars{\theta}\,\dd\theta}^{\ds{= 1}} \\[5mm] & = \bbx{{\root{2} \over 3}\,\pi} \approx 1.4810\\ & \end{align}

$\endgroup$
2
$\begingroup$

Your solid is precisely half the unit ball. Its volume is $\frac{2\pi}{3}$. One way to see it is to take the normal to the plane $x+2y+3z=0$, and rotate it to the vector $(0,0,1)$. The half-space $x+2y+3z\geq 0$ will then be rotated to the half-space $z\geq 0$, but the ball will remain unchanged, and it is evident that the intersection of $z\geq 0$ with the unit ball is half a ball.

$\endgroup$
3
  • 2
    $\begingroup$ I agree, and this is the way I would answer the question too, but OP might need to do an integration to stay in the spirit of the coursework. Hard to tell. $\endgroup$
    – Brian Tung
    Jun 8, 2021 at 22:50
  • $\begingroup$ Incidentally, one needn't rotate at all if one accepts that any plane passing through the origin must necessarily slice the unit sphere exactly in half. $\endgroup$
    – Brian Tung
    Jun 8, 2021 at 22:52
  • $\begingroup$ That is a very interesting insight I didn't see! However, I would much prefer to see a method using integration just for the sake of practice. Also, this may not hold with the edit I've added. The original problem was an ellipsoid with the constraint $x^2 + y^2+ 2z^2 \leq 1$. $\endgroup$ Jun 8, 2021 at 23:52
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Hereafter, $\ds{\bracks{\cdots}}$ is an $\ds{Iverson\ Bracket}$. \begin{align} V & \equiv \bbox[5px,#ffd]{\iiint_{\mathbb{R}^{3}} \bracks{x^{2} + y^{2} + 2z^{2} < 1}\bracks{x + 2y + 3z > 0} \dd x\,\dd y\,\dd z} \\[5mm] & = {\root{2} \over 2}\iiint_{\mathbb{R}^{3}} \bracks{x^{2} + y^{2} + z^{2} < 1} \bracks{x + 2y + {3\root{2} \over 2}\,z > 0}\dd x\,\dd y\,\dd z \\[5mm] & = \left.{\root{2} \over 2}\iiint_{r\ <\ 1} \bracks{\vec{r}\cdot\vec{n} > 0}\dd^{3}\vec{r} \,\right\vert_{\ \vec{\,\large n}\ \equiv\ \pars{1,2,3\root{2}/2}} = {\root{2} \over 2}\iiint_{r\ <\ 1} \int_{-\infty}^{\infty}{\expo{\ic k\,\vec{r}\cdot\vec{n}} \over k - \ic 0^{+}}\,{\dd k \over 2\pi\ic}\,\dd^{3}\vec{r} \\[5mm] & = {\root{2} \over 2}\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}}\iiint_{r\ <\ 1}\expo{\ic k\,\vec{r}\cdot\vec{n}} \,\dd^{3}\vec{r}\,{\dd k \over 2\pi\ic} \\[5mm] & = {\root{2} \over 2}\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}}\int_{0}^{1}4\pi r^{2} \underbrace{\int_{\Omega_{\,\vec{r}}}\expo{\ic k\,\vec{r}\cdot\vec{n}} \,{\dd\Omega_{\vec{r}} \over 4\pi}}_{\ds{\sin\pars{knr} \over knr}} \,\dd r\,{\dd k \over 2\pi\ic} \\[5mm] & = 2\root{2}\pi\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}} {1 \over \pars{kn}^{3}}\int_{0}^{kn}r\sin\pars{r}\,\dd r\,{\dd k \over 2\pi\ic} \\[5mm] & = 2\root{2}\pi\int_{-\infty}^{\infty}{1 \over k - \ic 0^{+}} {\sin\pars{kn} - kn\cos\pars{kn} \over \pars{kn}^{3}}\,{\dd k \over 2\pi\ic} \\[5mm] & = 2\root{2}\pi\,\ \underbrace{\lim_{k \to 0} \braces{\ic\pi\,{\sin\pars{kn} - kn\cos\pars{kn} \over \pars{kn}^{3}}\,{1 \over 2\pi\ic}}}_{\ds{1 \over 6}} = \bbx{{\root{2} \over 3}\,\pi} \approx 1.4810 \end{align}

$\endgroup$
2
  • $\begingroup$ I am unable to comprehend what you did here, but the answer seems nice and close to what I would expect from the unit ball. This problem is intended to be handled during a test without calculators, and with knowledge equivalent to someone who has taken basic analysis-based calculus courses. $\endgroup$ Jun 10, 2021 at 8:27
  • $\begingroup$ @Important_man74 I did it in that way because I wanted to use the ${\tt Heaviside\ Step\ Function\ Integral\ Representation}$ which indeed is not necessary. Anyway, I just added a shorter version. $\endgroup$ Jun 13, 2021 at 22:50

You must log in to answer this question.

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