[Math] How to compute the area of the portion of a paraboloid cut off by a plane

multiple integralmultivariable-calculussurface-integrals

How to compute:

The area of that portion of the paraboloid $x^2+z^2=2ay$ which is cut off by the plane $y=a$ ?

I think I have to compute $\iint f(x,z) dx dz$ , where $f(x,z)=\sqrt{(x^2+z^2)/2a}$ ; but I can't figure out what should be the limits of integration . Please help . Thanks in advance .

Best Answer

The problem is equivalent to find the volume of the solid bounded from above by $z=a$ and bounded from below by the surface $x^2+y^2=2az$. So the volume $V$ is the volume of the cylinder with base as $x^2+y^2=2a^2$ and altitude $z=a$ minus the volume under $x^2+y^2=2az$ and above the disc $x^2+y^2\leq2a^2$. Thus:

$$V=2a^2\pi a-\int_0^{2\pi}\int_0^{\sqrt2a}\frac{r^2}{2a}rdrd\theta=2a^3\pi-\frac{2\pi}{2a}\frac144a^4=\pi a^3.$$

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[Math] How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$

In order to calculate the surface of the cone $\mathcal{C}$ \begin{align*} \mathcal{C}:x^2+y^2=z^2 \end{align*} between the planes \begin{align*} z=0\qquad\text{ and }\qquad z=\frac{3-x}{2}\tag{1} \end{align*} we consider a parameter representation $\Phi(t,\varphi)$ of $\mathcal{C}$ \begin{align*} \Phi(t,\varphi) =\begin{pmatrix} x\\ y\\ z\end{pmatrix} =\begin{pmatrix} t\cos \varphi \\ t \sin \varphi\\ t \end{pmatrix}\qquad\qquad \begin{matrix}0\leq \varphi \leq 2\pi\\0\leq t \leq \frac{3}{2+\cos\varphi}\end{matrix}\tag{2} \end{align*}

Comment:

The apex of the cone is $(0,0,0)$ and at height $z=t$ the cone admits a representation by a circle with radius $t$ and polar coordinates: $(t\cos \varphi,t\sin \varphi)$.
The limits of the parameter representation $t$ in (2) are due to the planes in (1) \begin{align*} 0&\leq t\leq \frac{3-x}{2}\qquad\text{and}\qquad x=t\cos\varphi \end{align*}

The lateral surface $S_{lat}(\mathcal{C})$

The area $S_{lat}(\mathcal{C})$ of the lateral surface of $\mathcal{C}$ is \begin{align*} S_{lat}(\mathcal{C})&=\iint_\mathcal{C}\left\|\frac{\partial \Phi}{\partial \varphi}\times\frac{\partial\Phi}{\partial t}\right\|dS\\ &=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}} \left\|\begin{pmatrix}-t\sin\varphi\\t\cos\varphi\\0\end{pmatrix}\times\begin{pmatrix}\cos\varphi\\\sin\varphi\\1\end{pmatrix}\right\|\,dt\,d\varphi\\ &=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}} \left\|\begin{pmatrix}t\cos\varphi\\t\sin\varphi\\-t\end{pmatrix}\right\|\,dt\,d\varphi\\ &=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}} t\sqrt{2}\,dt\,d\varphi\\ &=\frac{9\sqrt{2}}{2}\int_{0}^{2\pi}\frac{1}{(2+\cos\varphi)^2} \,d\varphi\tag{3}\\ &=2\sqrt{6}\pi \end{align*}

The integral (3) was calculated with the help of WolframAlpha.

The top surface $S_{top}(\mathcal{C})$

The area of the top surface $S_{top}(\mathcal{C})$ is the intersection of the cone $\mathcal{C}$ with the plane $z=\frac{3-x}{2}$. Its projection on the $xy$-plane is the ellipse $\mathcal{E}:x^2+y^2=\left(\frac{3-x}{2}\right)^2$ and after normalisation we obtain \begin{align*} \mathcal{E}: \frac{1}{4}(x+1)^2+\frac{1}{3}y^2&=1 \end{align*} Let $\mathcal{D}$ denote the region within the ellipse $\mathcal{E}$ which has area $A(\mathcal{D})=2\sqrt{3}\pi$.

The area of the intersection of the plane $z=f(x,y)=\frac{3-x}{2}$ with the cone $\mathcal{C}$ is given by \begin{align*} S_{top}(\mathcal{C})&=\iint_\mathcal{D}\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}dA\\ &=\iint_\mathcal{D}\sqrt{\left(-\frac{1}{2}\right)^2+\left(0\right)^2+1}dA\\ &=\frac{\sqrt{5}}{2}\iint_\mathcal{D}dA\\ &=\frac{\sqrt{5}}{2}2\sqrt{3}\pi\\ &=\sqrt{15}\pi \end{align*}

$$ $$

We conclude: The area of the surface of the cone between the planes in (1) is \begin{align*} S_{lat}(\mathcal{C})+S_{top}(\mathcal{C})&=(2\sqrt{6}+\sqrt{15})\pi \end{align*}

[Math] To find area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$

First, lets see where both surfaces intersect. In polar coordinates, we have $$ z=3-r = 1+ r^2\quad \Rightarrow r=1. $$ So the intersection is the circle $r=1$ at height $z=2$.

The area of your surface is the area $A_1$ of $z=3-\sqrt{x^2+y^2}$ above this circle (let $S_1$ be this surface), plus the area $A_2$ of $z=1+x^2+y^2$ beneath this circle (let $S_2$ be this surface).

You can parametrize $S_1$ with the vectorial function $f$: $$ x=x, \quad y=y, \quad z=3-\sqrt{x^2+y^2}, $$ and $S_2$ with $g$: $$ x=x, \quad y=y, \quad z=3+x^2+y^2, $$ with $ (x,y)\in D=\{(r,\theta)\;|\;0\le r\le 1, 0\le \theta \le 2\pi\}$. Therefore, the total area equals $$ A_1+A_2=\iint_D ||f_x\times f_y|| dA + \iint_D ||g_x\times g_y|| dA\\ =\sqrt{2}\int_0^{2\pi}\int_{0}^1rdrd\theta+\int_0^{2\pi}\int_{0}^1r\sqrt{4r^2+1}drd\theta\\ =\sqrt{2}\pi+\frac{5\sqrt{5}-1}{6}\pi $$

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Related Solutions

[Math] How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$

[Math] To find area of the surface of the solid bounded by the cone $z=3-\sqrt{x^2+y^2}$ and the paraboloid $z=1+x^2+y^2$

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