Finding the volume of the region bounded by $z=\sqrt{\frac{x^2}{4}+y^2}$and $x+4z=a$. Cylindrical coordinates.

Question

I would like the answer to preferably be done using either using a surface integral, or an integral with substitutions. But anything other than this is alright, if nothing else exists.

I have to find the volume of the region bounded by $z=\sqrt{\frac{x^2}{4}+y^2}$and $x+4z=a$.

So, here we have a cone and a plane "cutting" it. I definitely must do this using some some of coordinate substitution. When doing a problem in class, that is, the area to be found being bounded in between $(z-1)^2=x^2+y^2$ and $z=0.$ the substitution was made, (which would be logical here to do aswell): $$x=r\cos\varphi \\ y=r \sin\varphi \\ z=z$$ and I also understand that the bounderies being $0\leq r\leq 1,0\leq\varphi\leq2\pi,0\leq z\leq 1-r.$ But in the problem I give that is a lot more difficult to do, I know that the bounderies for $\varphi$ should be the same,but with $z$ and $r$ I find it impossible. Should I find these conditional extreme points on the cone with the plane equation or am I not seeing something quite obvious here?

Answer 1

My solution is another way of stating @Jack D'Aurizio's solution. We want to rotate coordinates so that the plane $x+4z=a$ is horizontal. This can be achieved through $$\begin{array}{rl}x^{\prime}&=\frac4{\sqrt{17}}x-\frac1{\sqrt{17}}z\\ z^{\prime}&=\frac1{\sqrt{17}}x+\frac4{\sqrt{17}}z\end{array}$$ The inverse transformation is $$\begin{array}{rl}x&=\frac4{\sqrt{17}}x^{\prime}+\frac1{\sqrt{17}}z^{\prime}\\ z&=-\frac1{\sqrt{17}}x^{\prime}+\frac4{\sqrt{17}}z^{\prime}\end{array}$$ In terms of these coordinates, the plane is $z^{\prime}=\frac a{\sqrt{17}}$. Then the equation of the boundary reads $$\frac{(-x^{\prime}+4z^{\prime})^2}{17}=\frac{(4x^{\prime}+z^{\prime})^2}{4\cdot17}+y^2$$ Or $$(x^{\prime})^2-8x^{\prime}z^{\prime}+16(z^{\prime})^2=\frac14\left(16(x^{\prime})^2+8x^{\prime}z^{\prime}+(z^{\prime})^2\right)+17y^2$$ $$0=3(x^{\prime})^2-6x^{\prime}z^{\prime}-\frac{63}4(z^{\prime})^2+17y^2=3(x^{\prime}-z^{\prime})^2-\frac{75}4(z^{\prime})^2+17y^2$$ EDIT: Whoops, mistake on the above line. Should be $$0=3(x^{\prime})^2+10x^{\prime}z^{\prime}-\frac{63}4(z^{\prime})^2+17y^2=3\left(x^{\prime}+\frac53z^{\prime}\right)^2-\frac{289}{12}(z^{\prime})^2+17y^2$$ Now we can write this as the equation of an ellipse $$\frac{\left(x^{\prime}+\frac53z^{\prime}\right)^2}{\frac{289}{36}(z^{\prime})^2}+\frac{y^2}{\frac{17}{12}(z^{\prime})^2}=\frac{\left(x^{\prime}+\frac53z^{\prime}\right)^2}{\bar a^2}+\frac{y^2}{\bar b^2}=1$$ This is the equation of an ellipse with semi-major axis $\bar a=\frac{17}6z^{\prime}$, semi-minor axis $\bar b=\frac12\sqrt{\frac{17}3}z^{\prime}$ parallel to the $x^{\prime}y^{\prime}$-plane and centered at $\langle x_c^{\prime},y_c,z_c^{\prime}\rangle=\langle-\frac53z^{\prime},0,z^{\prime}\rangle$, so we know its area is $A(z^{\prime})=\pi\bar a\bar b=\pi\frac{17}{12}\sqrt{\frac{17}3}(z^{\prime})^2$, so we can add up the areas of the ellipses to get the volume $$V=\int_0^{\frac a{\sqrt{17}}}A(z^{\prime})dz^{\prime}=\int_0^{\frac a{\sqrt{17}}}\pi\frac{17}{12}\sqrt{\frac{17}3}(z^{\prime})^2dz^{\prime}=\left.\pi\frac{17}{12}\sqrt{\frac{17}3}\frac{(z^{\prime})^3}3\right|_0^{\frac a{\sqrt{17}}}=\frac{\pi a^3}{36\sqrt3}$$ So with the edit, my solution is now consistent with the others.

Answer 2

I think it is best to use some elementary geometry. The volume of a cone is one third of the product between the base area and the height, by Cavalieri's principle. The distance of the vertex of the cone (i.e. the origin) from the plane $x+4z=a$ is $\frac{a}{\sqrt{17}}$ by a well-known formula. The boundary of the base is the set of points $(x,y,z)$ such that $x+4z=a$ and $z^2=\frac{x^2}{4}+y^2$, that is a subset of the elliptic cylinder $\left(\frac{a-x}{4}\right)^2=\frac{x^2}{4}+y^2$. You just have to find the cross-section of such a cylinder (that depends on the determinant of the matrix associated with the last quadratic form) and the cosine of the angle between the plane $z=0$ and the plane $x+4z=0$ (easy to do with dot products) to have the base area.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{\mathcal{V} = \braces{\pars{x,y,z}\ \left.\vphantom{A^A}\right\vert\ \root{{x^{2} \over 4} + y^{2}}\ <\ z\ <\ {a - x \over 4}}}$:

\begin{align} \color{#f00}{\iiint_{\mathcal{V}}\dd x\,\dd y\,\dd z} & = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{z - \root{{x^{2} \over 4} + y^{2}}}\Theta\pars{{a - x \over 4} - z} \,\dd x\,\dd y\,\dd z \end{align}

Where $\Theta$ is the Heaviside Step function. Then,

\begin{align} \color{#f00}{\iiint_{\mathcal{V}}\dd x\,\dd y\,\dd z} & = 2\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \Theta\pars{z - \root{x^{2} + y^{2}}}\Theta\pars{{a - 2x \over 4} - z} \,\dd x\,\dd y\,\dd z \\[3mm] & = 2\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty} \Theta\pars{z - \rho}\Theta\pars{{a - 2\rho\cos\pars{\phi} \over 4} - z} \rho\,\dd\rho\,\dd\phi\,\dd z \\[3mm] & = 2\int_{0}^{2\pi}\int_{0}^{\infty} \bracks{{a - 2\rho\cos\pars{\phi} \over 4} - \rho} \Theta\pars{{a - 2\rho\cos\pars{\phi} \over 4} - \rho}\rho\,\dd\rho\,\dd\phi \\[3mm] & = \color{#88f}{2\int_{0}^{2\pi}\int_{0}^{\infty} \braces{{1 \over 4}\,a\rho - \half\bracks{2 + \cos\pars{\phi}}\rho^{2}} \Theta\pars{{a/2 \over 2 + \cos\pars{\phi}} - \rho}\,\dd\rho\,\dd\phi} \\[3mm] & = 2\int_{0}^{2\pi} \braces{{1 \over 8}\,a\,\bracks{a/2 \over 2 + \cos\pars{\phi}}^{2} - \half\bracks{2 + \cos\pars{\phi}} {1 \over 3}\bracks{{a/2 \over 2 + \cos\pars{\phi}}}^{3}}\,\dd\phi \\[3mm] & = {1 \over 48}\,a^{3}\ \overbrace{\int_{0}^{2\pi}{\dd\phi \over \bracks{2 + \cos\pars{\phi}}^{\,2}}} ^{\ds{{4\root{3} \over 9}\,\pi}}\ =\ \color{#f00}{{\root{3} \over 108}\,\pi a^{3}} \end{align}

Note that the $\color{#88f}{\mbox{"blue integration"}}$ requires $a > 0$ as we can see from the $\Theta$ argument. Otherwise it vanishes out.

Answer 4

Cylindrical coordinates is polar coordinates but in 3D. First you have to find projection in XY Plane which can be done so by eliminating z from above equations. Doing so you will get $$(x-\frac{a}{3})^2 - (\frac{4y}{\sqrt 3})^2 = (\frac{2a}{3})^2$$

Also your Z limits will be $ \sqrt{\frac{r^2cos^2\theta}{4}) + r^2sin^2\theta}$ and $\frac{a-rcos\theta}{4}$

To solve double integral you need to do substitution by putting $x=\frac{a}{3} + \frac{2a}{3}sec\theta$

$y= \frac{\sqrt3}{4}\frac{2a}{3}tan\theta$