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 $x+2z=3$?

Question

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 $x+2z=3$ ? I can't even figure out what the integrand will be ( should it be $\sqrt{z^2-x^2}$ ? ) and not even the limits . Please help . thanks in advance

Answer 1

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*}

Answer 2

The slope of the plane $x+2z=3$ is smaller than the slope of the cone, so their intersection curve is an ellipse, drawn with brown in the figure. The plane $z=0$ passes through the apex. Therefore, the region we are interested in is a ''hat''.

The projection of the surface region onto to the $x-y$ co-ordinate plane is another ellipse (the red one).

Between the two planes we have $0\le z\le \frac{3-x}2$, so the red ellipse is detemined by $$ x^2+y^2\le\left(\frac{3-x}2\right)^2 \\ \frac{(x+1)^2}4 + \frac{y^2}3 \le 1. $$ So, the area of the red ellipse is $2\sqrt3\pi$.

The slope of the cone is $1$, so the are of the two areas is $\sqrt2$. Therefore, the area of the conic region is $\sqrt2\cdot 2\sqrt3\pi=2\sqrt6\pi$.

(The question did not involve the area of the brown ellipse; just in order to verify the previous answer, from the slope of the plane $z=\frac{3-x}2$ it is $\sqrt{1^2+(\frac12)^2}\cdot 2\sqrt3\pi=\sqrt{15}\pi$.)