3.4 Surface Area

$S$ is a very simple geometric object.

$ab\sqrt{1+\tan^2\theta}=ab\sec\theta$

(a) $S$ is the part of the plane $z=y\,\tan\theta$ that lies above the rectangle in the $xy$-plane with vertices $(0,0)$, $(a,0)$, $(0,b)$, $(a,b)$. So $S$ is the rectangle with vertices $(0,0,0)$, $(a,0,0)$, $(0,b,b\tan\theta)$, $(a,b,b\tan\theta)$. So it has side lengths

and hence area $ab\sqrt{1+\tan^2\theta}=ab\sec\theta$.

(b) $S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = y\,\tan\theta$ and with $(x,y)$ running over

Hence by Theorem 3.4.2 in the CLP-3 text

$\frac{\sqrt{a^2+b^2+c^2}}{c} A(D)$

$S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = \frac{d-ax-by}{c}$ and with $(x,y)$ running over $D$. Hence by Theorem 3.4.2 in the CLP-3 text

The triangle is part of the plane $\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1$.

(a) $\frac{1}{2}\sqrt{a^2b^2+a^2c^2+b^2c^2}$

(b) See the solution.

Note that all three vertices $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ lie on the plane $\frac{x}{a}+\frac{y}{b} +\frac{z}{c}=1$. So the triangle is part of that plane.

Method 1.
$S$ is the part of the surface $z=f(x,y)$ with $f(x,y) = c\left(1-\frac{x}{a}-\frac{y}{b}\right)$ and with $(x,y)$ running over the triangle $T_{xy}$ in the $xy$-plane with vertices $(0,0,0)$ $(a,0,0)$ and $(0,b,0)$. Hence by part a of Theorem 3.4.2 in the CLP-3 text

where $A(T_{xy})$ is the area of $T_{xy}$. Since the triangle $T_{xy}$ has base $a$ and height $b$ (see the figure below), it has area $\frac{1}{2}ab$. So

Method 2.
$S$ is the part of the surface $x=g(y,z)$ with $g(y,z) = a\left(1-\frac{y}{b}-\frac{z}{c}\right)$ and with $(y,z)$ running over the triangle $T_{yz}$ in the $yz$-plane with vertices $(0,0,0)$ $(0,b,0)$ and $(0,0,c)$. Hence by part b of Theorem 3.4.2 in the CLP-3 text

where $A(T_{yz})$ is the area of $T_{yz}$. Since $T_{yz}$ has base $b$ and height $c$, it has area $\frac{1}{2}bc$. So

Method 3.
$S$ is the part of the surface $y=h(x,z)$ with $h(x,z) = b\left(1-\frac{x}{a}-\frac{z}{c}\right)$ and with $(x,z)$ running over the triangle $T_{xz}$ in the $xz$-plane with vertices $(0,0,0)$ $(a,0,0)$ and $(0,0,c)$. Hence by part c of Theorem 3.4.2 in the CLP-3 text

where $A(T_{xz})$ is the area of $T_{xz}$. Since $T_{xz}$ has base $a$ and height $c$, it has area $\frac{1}{2}ac$. So

(b) We have already seen in the solution to part (a) that

Hence

$\frac{8}{27}\left[\left(\frac{13}{4}\right)^{3/2}-1\right]$

For the surface $z=f(x,y)=y^{3/2}$,

by Theorem 3.4.2.a in the CLP-3 text, So the area is

$\frac{\pi}{6}\big[{(1+4a^2)}^{3/2}-1\big]$

First observe that any point $(x,y,z)$ on the paraboliod lies above the $xy$-plane if and only if

That is, if and only if $(x,y)$ lies in the circular disk of radius $a$ centred on the origin. The equation of the paraboloid is of the form $z=f(x,y)$ with $f(x,y)=a^2-x^2-y^2$. So, by Theorem 3.4.2.a in the CLP-3 text,

Switching to polar coordinates,

$5\sqrt{2}\pi$

First observe that any point $(x,y,z)$ on the cone lies between the planes $z=2$ and $z=3$ if and only if $4\le x^2+y^2\le 9$.

The equation of the cone can be rewritten in the form $z=f(x,y)$ with $f(x,y)=\sqrt{x^2+y^2}$. Note that

So, by Theorem 3.4.2.a in the CLP-3 text,

Now the domain of integration is a circular washer with outside radius $3$ and inside radius $2$ and hence of area $\pi(3^2-2^2)=5\pi$. So the surface area is $5\sqrt{2}\pi$.

$\frac{4}{15}\big[9\sqrt{3}-8\sqrt{2}+1\big]$

The equation of the surface is of the form $z=f(x,y)$ with $f(x,y)=\frac{2}{3}\big(x^{3/2} + y^{3/2}\big)$. Note that

So, by Theorem 3.4.2.a in the CLP-3 text,

The total surface area of (b) (ii) can be determined without evaluating any integrals.

(a) $F(x,y) = \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}$ (b) (i) $\int_0^{2\pi}\dee{\theta}\int_0^1\dee{r}\ \frac{2r}{\sqrt{4-r^2}}$ (ii) $16\pi$

(a) By Theorem 3.4.2.a in the CLP-3 text, $F(x,y) = \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}$.

(b) (i) The “dimple” to be painted is part of the upper sphere $x^2+y^2+\big(z-2\sqrt{3}\big)^2=4$. It is on the bottom half of the sphere and so has equation $z=f(x,y)=2\sqrt{3}-\sqrt{4-x^2-y^2}$. Note that

The point on the dimple with the largest value of $x$ is $(1,0,\sqrt{3})$. (It is marked by a dot in the figure above.) The dimple is invariant under rotations around the $z$–axis and so has $(x,y)$ running over $x^2+y^2\le 1$. So, by Theorem 3.4.2.a in the CLP-3 text,

Switching to polar coordinates,

(b) (ii) Observe that if we flip the dimple up by reflecting it in the plane $z=\sqrt{3}$, as in the figure below, the “Death Star” becomes a perfect ball of radius $2$.

The area of the pink dimple in the figure above is identical to the area of the blue cap in that figure. So the total surface area of the Death Star is exactly the surface area of a sphere of radius $a=2$ and so (see Example 3.4.5 in the CLP-3 text) is $4\pi a^2 = 4 \pi 2^2=16\pi$.

Rewrite the equation of the cone in the form $x=g(y,z)$.

$6\pi\approx 18.85$

The equation of the half of the cone with $x\ge 0$ can be rewritten in the form $x=g(y,z)$ with $g(y,z)=\sqrt{3}\sqrt{y^2+z^2}$. Note that

so that, by Theorem 3.4.2 in the CLP-3 text,

A point $(x,y,z)$ on $x=\sqrt{3}\sqrt{y^2+z^2}$ has $\sqrt{3}\le x\le 2\sqrt{3}$ if and only if $1\le\sqrt{y^2+z^2}\le 2$. So

$a^2[\pi-2]$

We are to find the surface area of part of a hemisphere. On the hemisphere

so that

In polar coordinates, this is $\dee{S}=\frac{a}{\sqrt{a^2-r^2}}\,r\,\dee{r}\,\dee{\theta}$. We are to find the surface area of the part of the hemisphere that is inside the cylinder, $x^2-ax+y^2=0$, which is polar coordinates is becomes $r^2-ar\cos\theta=0$ or $r=a\cos\theta$. The top half of the domain of integration is sketched below.

So the