\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)
The portion of the cone \(\displaystyle z = \sqrt{x^2 + y^2}\) below the plane \(\displaystyle z = 4\)
So I have to find the slanted surface area of the cone.
\(\displaystyle S_A = \iint_{S} |d\bold{S}| = \iint_{S} |\bold{n}| \ dS = \iint_{S} dS = \iint_{D} |\bold{m}| \ dx \ dy\)
\(\displaystyle \bold{r}(x,y) = \left(x,y,\sqrt{x^2 + y^2}\right)\)
\(\displaystyle \bold{m} = \bold{r}_x \times \bold{r}_y\)
\(\displaystyle \bold{r}_x = \left(1,0,\frac{x}{\sqrt{x^2 + y^2}}\right)\)
\(\displaystyle \bold{r}_y = \left(0,1,\frac{y}{\sqrt{x^2 + y^2}}\right)\)
\(\displaystyle \bold{r}_x \times \bold{r}_y = \begin{vmatrix} \bold{i} & \bold{j} & \bold{k} \\ 1 & 0 & \dfrac{x}{\sqrt{x^2 + y^2}} \\ 0 & 1 & \dfrac{y}{\sqrt{x^2 + y^2}} \end{vmatrix} = \left( -\frac{x}{\sqrt{x^2 + y^2}},\ -\frac{y}{\sqrt{x^2 + y^2}},\ 1 \right) \)
Then,
\(\displaystyle |\bold{m}| = \left| \bold{r}_x \times \bold{r}_y \right| = \sqrt{2} \)
This gives:
\(\displaystyle S_A = \iint_{D} |\bold{m}| \ dx \ dy = \iint_{D} \sqrt{2} \ dx \ dy\)
\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)
The portion of the cone \(\displaystyle z = \sqrt{x^2 + y^2}\) below the plane \(\displaystyle z = 4\)
Substitute \(\displaystyle z = 4\) in \(\displaystyle z = \sqrt{x^2 + y^2}\)
\(\displaystyle 4 = \sqrt{x^2 + y^2}\)
Or
\(\displaystyle 16 = x^2 + y^2\)
Or
\(\displaystyle 4^2 = x^2 + y^2\)
This is the region on the \(\displaystyle xy\)-plane which is just a circle of radius \(\displaystyle 4\).
Then,
\(\displaystyle S_A = \iint_{D} \sqrt{2} \ dx \ dy = \sqrt{2}\int_{-4}^{4}\int_{-\sqrt{16 - y^2}}^{\sqrt{16 - y^2}} \ dx \ dy\)
Or easier to work in the polar coordinate.
\(\displaystyle S_A = \sqrt{2}\int_{-4}^{4}\int_{-\sqrt{16 - y^2}}^{\sqrt{16 - y^2}} \ dx \ dy = \sqrt{2}\int_{0}^{2\pi}\int_{0}^{4} r \ dr \ d\theta = \textcolor{blue}{16\pi\sqrt{2}}\)
