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 $x+2z=3$? 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 
 A: 
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*}

A: 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 $\sqrt{1^2+(\frac12)^2}\cdot 2\sqrt3\pi=\sqrt{15}\pi$.)
