
\section{Area Integrals}

\begin{enumerate}[1.]

\item
{\it Evaluate the surface integral
$\iint_{\mathcal S}\sqrt{x} \, dS$, if 
$\mathcal{S}$ is the parametrized surface ${\bf r}(u,v)=(u^2,v,u+v )$ with $0\leq u\leq 1$, $0\leq v\leq 1$.}

\medskip


We have ${\bf r}_u = (2u,0,1),\ \ {\bf r}_v = (0,1,1)$, so
$$
{\bf n}={\bf r}_u\times {\bf r}_v=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
2u& 0& 1 \\
0& 1& 1\\
            \end{matrix}\right|
=- {\bf i} -2 u {\bf j} + 2u{\bf k},
$$
$$|{\bf n}|=\sqrt{1+4u^2 + 4u^2 } = \sqrt{1+8u^2}.
$$
Therefore, with the substitution $8u^2+1 =w$, $16u du = dw$,
\begin{align*}
\iint_{\mathcal S} \sqrt{x}dS
&= \int_0^1 \int_0^1  u \sqrt{1+8u^2} \, dv du 
\\
&= \int_0^1   u \sqrt{1+8u^2} du
\\
&= \int_1^9  \sqrt{w} \, \frac{dw}{16}  
\\
&= \frac{w^{3/2}}{(3/2)\cdot 16}\Big|_1^9 = \frac{27-1}{24}=\frac{26}{24}=\frac{13}{12}.
\end{align*}

\bigskip

\item
{\it A parametric surface is given by the equations $x=uv, y=u^2-v^2, z=u^2+v^2$, with $u,v\geq 0$.
Find the equation of the plane tangent to the surface at the point  
$P$ where $(u,v)=(2,1)$. }

\medskip

We have
$\bfr_u=\langle v,2u,2u \rangle$ and $\bfr_v=\langle u,-2v,2v \rangle$, so that the normal vector is
$${\bf n}=\bfr_u\times \bfr_v=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
v& 2u& 2u \\
u & -2v & 2v  \\
            \end{matrix}\right|
 =8uv \ii - (2v^2-2u^2) \jj + (-2v^2-2u^2) \kk \ . $$
The point $P$ has coordinates $(2, 3, 5)$, and the normal vector at $P$ is
${\bf n}= 16 \ii +6 \jj -10\kk$.  
Therefore the tangent plane is
$$
16 (x-2 ) +6 (y-3 )-10 (z-5)=0,
$$
which can be simplified to $8(x-2)+3(y-3)-5 (z-5)=0$, $8x+3y-5z=0$.

 \bigskip

\item
{\it Let $\mathcal{S}$ be the surface ${\bf r}(u,v)=(u \cos v, u\sin v, v)$ with $0\leq u\leq 1$, $0\leq v\leq \pi$.

\medskip
\noindent
(a) Find the tangent plane to the surface at the point where $u=1/2,v=\pi/6$.}

\medskip

We have ${\bf r}_u=\cos v \ii + \sin v \jj$ and ${\bf r}_v=-u\sin v \ii + u\cos v \jj +\kk$, so that
$${\bf n}=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
\cos v& \sin v& 0\\
-u \sin v& u\cos v& 1\\
            \end{matrix}\right|
 =\sin v \ii -\cos v\jj + u\kk
 $$
At $(u,v)=(1/2,\pi/6)$, 
$${\bf r}=\left\langle \frac{1}{2}\cos\frac{\pi}{6}, \frac{1}{2}\sin \frac{\pi}{6}, \frac{\pi}{6}\right\rangle= 
\left\langle \frac{\sqrt{3}}{4},  \frac{1}{4}, \frac{\pi}{6}\right\rangle ,$$
$${\bf n}= \left\langle \sin \frac{\pi}{6}, -\cos  \frac{\pi}{6}, \frac{1}{2}\right\rangle= 
\frac{1}{2}\langle1,-\sqrt{3},1\rangle,\ \ 
$$
so the equation of the tangent plane is
$$
\left(x-\frac{\sqrt{3}}{4}\right) - \sqrt{3}\left(y-\frac{1}{4}\right) + \left(z-\frac{\pi}{6}\right)=0,$$
which simplifies to $x-\sqrt{3}y+z=\pi/6$.




\bigskip
\noindent
{\it (b) Evaluate the surface integral $\iint_{\mathcal{S}} (x^2+y^2)^{1/2} dS$.}

\medskip

From (a), we have $|{\bf n}|=(\sin^2 v + \cos^2 v +u^2)^{1/2}=\sqrt{1+u^2}$. Now
 $$
 \iint_{\mathcal{S}} (x^2+y^2)^{1/2} dS
 =\int_0^\pi \int_0^1 u \sqrt{1+u^2}\, dudv
 $$
 We substitute $t=1+u^2$, $2udu = dt$, $u^2=t-1$, so that the last integral is
 $$
 \frac{1}{2}\int_0^\pi \int_1^2  \sqrt{t}\,  dt dv
 =\frac{1}{2}\int_0^\pi \frac{2}{3}t^{3/2} \Big|_1^2 dv
  =\frac{1}{2}\int_0^\pi  \frac{2}{3}(2^{3/2}-1) dv 
 =\frac{\pi}{3} (2^{3/2}-1)
 $$
 
 \bigskip 
 
\item
{\it   
Let $\mathcal{S}$ be the surface parametrized by ${\bf r}(u.v)=(u+\sin u,v+\sin v,1-\cos(u+v))$, $0<u<\pi$,
$0<v<\pi$ (the boundary of the rectangle is not included). 


\medskip
\noindent(a) Find all $(u,v)$ such that the tangent plane to the surface at ${\bf r}(u,v)$ is parallel
to the $xy$-plane. }


\bigskip


We calculate the normal vector:
\medskip
$${\bf r}_u=\langle 1+\cos u,0,\sin(u+v)\rangle,$$
$${\bf r}_v=\langle 0,1+\cos v,\sin(u+v)\rangle,$$
\begin{align*}
{\bf n}&={\bf r}_u\times {\bf r}_v
\\
&=-(1+\cos v)\sin(u+v)\ii -(1+\cos u)\sin(u+v)\jj
+(1+\cos u)(1+\cos v)\kk.
\end{align*}
For the tangent plane to be parallel to the $xy$-plane, ${\bf n}$ 
needs to be parallel to $\kk$, so its first two components should 
be 0:
$$(1+\cos v)\sin(u+v)=(1+\cos u)\sin(u+v)=0.$$
Since $1+\cos u$ and $1+\cos v$ are nonzero for $u,v$ in $(0,\pi)$
(remember that we are excluding the boundary), we must have 
$\sin(u+v)=0$, $u+v=\pi$.  The corresponding points on the surface
are 
$${\bf r}(u,\pi-u)=(u+\sin u,\pi-u+\sin u, 2),\ 0<u<\pi.$$


\medskip
\noindent{\it (b) 
Prove that the area of $\mathcal{S}$ is less than $5\pi^2$. }



\medskip
The area is $\displaystyle{\int_0^\pi\int_0^\pi |{\bf n}(u,v)|dudv}$.
From part (a), we have
$$|{\bf n}|^2
=(1+\cos v)^2 \sin(u+v)^2 +(1+\cos u)^2 \sin(u+v)^2 
+(1+\cos u)^2(1+\cos v)^2.$$
Since sine and cosine are always between $-1$ and $1$, the first term
on the right is at most 4, the second term is at most 4, and the third
term is at most $4\cdot 4=16$.  Hence 
$|{\bf n}|\leq\sqrt{4+4+16}=\sqrt{24}$, so that 
$$\hbox{area}\leq \int_0^\pi\int_0^\pi \sqrt{24}\,dudv=\sqrt{24}\pi^2<5\pi^2.$$

\bigskip




\item {\it Evaluate the integral
$\iint_{\mathcal S} (x^2+4y^2+1) \, dS$, if 
$\mathcal{S}$ is the part of the paraboloid $z=\frac{x^2}{2}+y^2$ that lies above the region 
$D=\{(x,y):\ x^2+4y^2\leq 1$ and $y\geq 0\}$
in the $xy$-plane.}


\medskip

We parametrize the surface by ${\bf r}(x,y)= \langle x,y, \frac{x^2}{2}+y^2 \rangle$. Then
${\bf r}_x = \langle 1,0,x\rangle$ and ${\bf r}_y = \langle 0,1,2y\rangle$, so that
$$
{\bf n}={\bf r}_x\times {\bf r}_y=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
1 & 0 & x \\
0& 1&  2y \\
            \end{matrix}\right|
=-x {\bf i} -2 y {\bf j} + {\bf k},
$$
$$|{\bf n}|=\sqrt{x^2+4y^2 +1 } ,
$$
\begin{align*}
\iint_{\mathcal S} (x^2+4y^2+1) \, dS &= \iint_D (x^2+4y^2+1)  \sqrt{x^2+4y^2 +1 } dA
\\
&= \iint_D (x^2+4y^2+1)^{3/2}dA
\end{align*}
To evaluate this integral, we use modified polar coordinates $x=r\cos\theta$, $y=\frac{1}{2} r\sin\theta$,
$$
\frac{\partial(x,y)}{\partial(r,\theta)} = \left|\begin{matrix}
\cos\theta & \frac{1}{2}\sin\theta \\
-r\sin\theta & \frac{1}{2} r\cos\theta  \\
            \end{matrix}\right| = \frac{1}{2} r\cos^2 \theta + \frac{1}{2} r\sin^2\theta    =\frac{r}{2} 
$$
With that substitution, $x^2+4y^2= r^2 \cos^2 \theta + 4\cdot \frac{1}{4} r^2\sin^2\theta = r^2$,
so that
\begin{align*}
\iint_{\mathcal S} (x^2+4y^2+1) \, dS &= \int_0^1 \int_0^\pi \left(  r^2+1  \right)^{3/2}
 \frac{r}{2}        d\theta dr 
\\
&= \frac{\pi}{4} \int_0^1 (r^2+1)^{3/2} (2r) dr 
\\
&= \frac{\pi}{4} \frac{(r^2+1)^{5/2}}{5/2}  \Big|_{r=0}^{1}    
= \frac{\pi}{4} \frac{2^{5/2}-1}{5/2} = \frac{\pi (2^{5/2}-1)}{10} 
\end{align*}



\bigskip


\item {\it 
Let $\mathcal{C}$ be a smooth simple curve in the plane $z=0$, parametrized by ${\bf r}(t)=x(t)\ii+y(t)\jj$ with $a\leq t\leq b$. Construct a surface $\mathcal{S}$ in three dimensions by taking the union of all line segments that connect points ${\bf r}(t)$ on the curve to the fixed point $P=(0,0,1)$. Prove that
$$
\frac{1}{2}(\hbox{length of }\mathcal{C}) \leq  (\hbox{area of }\mathcal{S}).
$$
Give an example of $\mathcal{C}$ such that equality holds in the above formula.

(Hint: How would you parametrize the surface? You may want to start by parametrizing each line segment separately.)}

\medskip

First, the length of the curve is
\begin{equation}\label{e1}
L=\int_a^b |\bfr'(t)|dt= \int_a^b  \sqrt{(x')^2+(y')^2} dt
\\
\end{equation}
Next, we parametrize the surface by the vector-valued function ${\bf R}(t,s)$:
$$
{\bf R}(t,s)= (1-s)\kk + s\bfr(t),\ \ a\leq t\leq b,\ 0\leq s\leq 1.
$$
Using this parametrization, we have
$$
{\bf n}=
\bfr_t \times \bfr_s=
\left|\begin{matrix}
{\bf i} & {\bf j} & {\bf k}\\
sx'(t) & sy'(t) & 0\\
x(t) & y(t) & -1\\
            \end{matrix}\right|
= -sy' \ii + sx' \jj +(sx'y-sxy')\kk,
$$
so that
\begin{align*}
|{\bf n}|&=s\sqrt{(x')^2+(y')^2+(x'y-xy')^2}
\\
&\geq s\sqrt{(x')^2+(y')^2}.
\end{align*}
The area of the surface is
\begin{align*}
A&=\iint_\mathcal{S} dS=\int_a^b \int_0^1 s\sqrt{(x')^2+(y')^2+(x'y-xy')^2} dsdt
\\
&= \int_a^b \frac{s^2}{2}\Big|_{s=0}^1 \sqrt{(x')^2+(y')^2+(x'y-xy')^2} dt
\\
&= \frac{1}{2} \int_a^b  \sqrt{(x')^2+(y')^2+(x'y-xy')^2} dt
\\
&\geq \frac{1}{2} \int_a^b \sqrt{(x')^2+(y')^2} dt=\frac{1}{2}L,
\end{align*}
as required. 

For an example where the equality holds, let $\mathcal{C}$ be the line segment from $(0,0,0)$ to $(1,0,0)$. Then $\mathcal{C}$ has length 1. The surface $\mathcal{S}$ is the right triangle with vertices at $(0,0,0)$, $(1,0,0)$, and $(0,0,1)$, and has area 1/2, which is half of the length of $\mathcal{C}$.





\item {\it
Let $D$ be a planar domain such that 
$$
D\subset \{(x,y):\ x^2+y^2\leq 64\hbox{ and } \frac{x^2}{36}+\frac{y^2}{49}\geq 1\}.
$$
(This is just an inclusion and $D$ does not have to be equal to the set on the right.)

Let $\mathcal{S}$ be the part of the sphere $x^2+y^2+z^2=64$ that lies above $D$. Prove that
$$
(\hbox{area of }D)\leq\frac{3}{4} (\hbox{area of }\mathcal{S}).
$$

}

\medskip

We have 
$$
(\hbox{area of }\mathcal{S})= \iint_\mathcal{S}\,d{\bf S} 
= \iint_D\frac{|\nabla G|}{|G_z|}dxdy ,
$$
where $G(x,y,z)=x^2+y^2+z^2$. We have for $(x,y,z)\in\mathcal{S}$,
$$|\nabla G |=\sqrt{4x^2+4y^2+4z^2}= 2\sqrt{x^2+y^2+z^2}=16,$$
$$|G_z|=2z=2\sqrt{64-x^2-y^2}.$$
On $D$, we have 
$$\frac{x^2}{36}+\frac{y^2}{36}\geq 
\frac{x^2}{36}+\frac{y^2}{49}\geq 1,$$
so that $x^2+y^2\geq 36$ and $\sqrt{64-x^2-y^2}\leq \sqrt{64-36} = \sqrt{28}$.
Therefore
$$
\frac{|\nabla G|}{|G_z|}\geq \frac{16}{2\sqrt{28}}= \frac{4}{\sqrt{7}}>\frac{4}{3},
$$
$$
(\hbox{area of }\mathcal{S})> \iint_D\frac{4}{3}dxdy = \frac{4}{3}(\hbox{area of }D). 
$$



\item {\it (
Let $\mathcal{S}$ be the part of the sphere $x^2+y^2+z^2=1$ that lies above the cone $z=\sqrt{x^2+y^2}$. Consider the integral
$$
\iint_{\mathcal{S}} \sqrt{x^2+y^2} dS.
$$
A student has evaluated this integral as follows.



\medskip\noindent
\fbox{
\begin{minipage}{0.8\textwidth}

The unit normal to $\mathcal{S}$ is the vector ${\bf N}=x\ii +y\jj +z\kk$. The angle $\gamma$ between ${\bf N}$ and $\kk$ is $\gamma=\phi$ in spherical coordinates, so that the area element is $\frac{1}{\cos\gamma}=\frac{1}{\cos\phi}$. Also in spherical coordinates, $\sqrt{x^2+y^2}=\sin\phi$ on the sphere. Therefore
\begin{align*}
\iint_{\mathcal{S}} \sqrt{x^2+y^2} dS
&= \int_0^{2\pi} \int_0^{\pi/4} \frac{\sqrt{x^2+y^2}}{\cos\gamma } d\phi\, d\theta\\
&= \int_0^{2\pi} \int_0^{\pi/4} \frac{\sin\phi}{\cos\phi} d\phi\, d\theta\\
&= 2\pi \Big[ -\ln(\cos\phi) \Big]_{0}^{\pi/4}\\
&= 2\pi \Big( -  \ln(\cos(\pi/4))  + \ln(\cos 0)\Big) \\
& =2\pi (- \ln(\sqrt{2}/2)- \ln(1) ) = - 2\pi \ln(\sqrt{2}/2).
\end{align*}
\end{minipage}
}

Is this solution correct? If not, explain where the student has made a mistake (if there is more than one error, indicate all of them), and provide the correct solution.}



\bigskip


The mistake is that the area element is $dS=\frac{1}{\cos\gamma}dxdy =\frac{1}{\cos\phi}dxdy$ in {\it rectangular} coordinates. In spherical coordinates, we have $dS=\sin\phi d\phi d\theta$, and 

\begin{align*}
\iint_{\mathcal{S}} \sqrt{x^2+y^2} dS
&= \int_0^{2\pi} \int_0^{\pi/4} \sqrt{x^2+y^2}\sin\phi \,d\phi\, d\theta\\
&= \int_0^{2\pi} \int_0^{\pi/4} \sin^2 \phi\, d\phi\, d\theta\\
&= 2\pi  \int_0^{\pi/4} \frac{1-\cos(2\phi)}{2}\, d\phi
\\
&= \pi \Big[ \phi-\frac{\sin(2\phi)}{2}  \Big]_{0}^{\pi/4}\\
&= \pi \Big[ \frac{\pi}{4}-\frac{1}{2}  \Big].
\end{align*}







\end{enumerate}




