Parametrized Surfaces

Section 3.1 Parametrized Surfaces

For many applications we will need to use integrals over surfaces. One obvious one is just computing surface areas. Another is computing the rate at which fluid traverses a surface. The first step is to simply specify surfaces carefully.

There are three common ways to specify a surface in three dimensions.

Graph of a function: Probably the most common way to specify a surface is to give its equation in the form
\begin{equation*} z = f(x,y)\qquad (x,y)\in\cD\subset\bbbr^2 \end{equation*}
Here “$(x,y)\in\cD\subset\bbbr^2$” just means that $(x,y)$ runs over the subset $\cD$ of $\bbbr^2\text{.}$ For example, if the surface is the top half of the sphere of radius one centred on the origin
\begin{equation*} z = \sqrt{1-x^2-y^2}\qquad \text{with } x^2+y^2 \le 1 \end{equation*}
Implicitly: We can also specify that the surface is the set of points $(x,y,z)$ that satisfy the equation $G(x,y,z)=0\text{,}$ or, more generally
¹
Of course we can always convert the equation $G(x,y,z)=K$ into $H(x,y,z)=0$ with $H(x,y,z)=G(x,y,z)-K\text{.}$ But it is often more convenient to use $G(x,y,z)=K\text{.}$
, satisfy the equation $G(x,y,z)=K\text{,}$ with $K$ a constant. For example, the sphere of radius one centred on the origin is the set of points that obey
\begin{equation*} x^2+y^2+z^2=1 \end{equation*}
We shall explore this surface a little more in Example 3.1.2 below.
Range of a function: Probably the most useful way to specify a surface, when one needs to integrate over the surface, is as the range of a function
\begin{align*} \vr&: \cD\subset\bbbr^2 \rightarrow \bbbr^3\\ &(u,v) \in\cD \mapsto \vr(u,v) =\big(x(u,v)\,,\,y(u,v)\,,\,z(u,v)\big) \end{align*}
The upper line means that $\vr$ is a function which is defined on the subset $\cD$ of $\bbbr^2$ and which assigns to each point on $\cD$ a point in $\bbbr^3\text{.}$ The second line means that the function $\vr$ assigns to the element $(u,v)$ of $\cD$ the element $\vr(u,v) =\big(x(u,v)\,,\,y(u,v)\,,\,z(u,v)\big)$ in $\bbbr^3\text{.}$ Such a surface is called a parametrized surface — each point of the surface is labelled by the values of the two parameters $u$ and $v\text{.}$ Parametrized surfaces are of course the two parameter analog of parametrized curves. Examples of parametrized surfaces come next.

Example 3.1.1.

One simple, even trivial, way to parametrize the surface which is the graph

\begin{equation*} z = f(x,y)\qquad (x,y)\in\cD\subset\bbbr^2 \end{equation*}

is to choose $x$ and $y$ as the parameters. That is, to choose

\begin{align*} \vr(u,v) &= \big(u,v,\,f(u,v)\big),\quad (u,v)\in\cD\\ \text{or}\qquad \vr(x,y) &= \big(x,y,\,f(x,y)\big),\quad (x,y)\in\cD \end{align*}

Let’s do something a bit more substantial.

Example 3.1.2. Sphere.

The sphere of radius $1$ centred on the origin is the set of points $(x,y,z)$ that obey

\begin{equation*} G(x,y,z)= x^2+y^2+z^2=1 \end{equation*}

We cannot express this surface as the graph of a function because, for each $(x,y)$ with $x^2+y^2 \lt 1\text{,}$ there are two $z$’s that obey $x^2+y^2+z^2=1\text{,}$ namely

\begin{equation*} z=\pm\sqrt{1-x^2-y^2} \end{equation*}

On the other hand, locally, this surface is the graph of a function. This means that, for any point $(x_0, y_0, z_0)$ on the sphere, all points of the surface that are sufficiently near $(x_0, y_0, z_0)$ can be expressed in one of the forms $z=f(x,y)$ or $x=g(y,z)\text{,}$ or $y=h(x,z)\text{.}$ For example, the part of the sphere that is within a distance $\sqrt{2}$ of the point $(0,0,1)$ is

\begin{align*} &\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ |(x,y,z) - (0,0,1)| \lt \sqrt{2}}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ x^2+y^2+(z-1)^2 \lt 2}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ x^2+y^2+z^2-2z+1 \lt 2}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ z \gt 0}\\ &=\Set{(x,y,z)}{ z=\sqrt{1-x^2-y^2},\ x^2+y^2 \lt 1} \end{align*}

This is illustrated in the figure below which shows the $y=0$ section of the sphere $x^2+y^2+z^2=1$ and also the $y=0$ section of the set of points that are within a distance $\sqrt{2}$ of $(0,0,1)\text{.}$ (They are the points inside the dashed circle.)

Similarly, as illustrated schematically in the figure below, the part of the sphere that is within a distance $\sqrt{2}$ of the point $(1,0,0)$ is

\begin{align*} &\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ |(x,y,z) - (1,0,0)| \lt \sqrt{2}}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ (x-1)^2+y^2+z^2 \lt 2}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ x^2-2x+1+y^2+z^2 \lt 2}\\ &=\Set{(x,y,z)}{ x^2+y^2+z^2=1,\ x \gt 0}\\ &=\Set{(x,y,z)}{ x=\sqrt{1-y^2-z^2},\ y^2+z^2 \lt 1} \end{align*}

The figure below shows the $y=0$ section of the sphere $x^2+y^2+z^2=1$ and also the $y=0$ section of the set of points that are within a distance $\sqrt{2}$ of $(1,0,0)\text{.}$ (Again, they are the points inside the dashed circle.)

We can parametrize the unit sphere by using spherical coordinates, which you should have seen before. As a reminder, here is a figure showing the definitions of the three spherical coordinates

The symbols $\rho\text{,}$ $\varphi\text{,}$ $\theta$ are the standard mathematics symbols for the spherical coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering.

\begin{align*} \rho&=\text{ distance from }(x,y,z)\text{ to }(0,0,0)\\ \varphi&=\text{ angle between the line }\overline{(0,0,0)\,(x,y,z)} \text{ and the $z$ axis}\\ \theta&=\text{ angle between the line }\overline{(0,0,0)\,(x,y,0)} \text{ and the $x$ axis} \end{align*}

and here are two more figures giving the side and top views of the previous figure.

From the figure, we see that Cartesian and spherical coordinates are related by

\begin{align*} x&=\rho\sin\varphi\cos\theta\\ y&=\rho\sin\varphi\sin\theta\\ z&=\rho\cos\varphi \end{align*}

The points on the sphere $x^2+y^2+z^2=1$ are precisely the set of points with $\rho=1\text{.}$ So we can use the parametrization

\begin{equation*} \vr(\theta,\varphi) =\big(\sin\varphi\cos\theta\,,\sin\varphi\sin\theta\,,\,\cos\varphi\big) \end{equation*}

Here is how to see that as $\varphi$ runs over $(0,\pi)$ and $\theta$ runs over $[0,2\pi)\text{,}$ $\vr(\theta,\varphi)$ covers the whole sphere $x^2+y^2+z^2=1$ except for the north pole ($\varphi=0$ gives the north pole for all values of $\theta$) and the south pole ($\varphi=\pi$ gives the south pole for all values of $\theta$).

Fix $\theta$ and have $\varphi$ run over the interval $0 \lt \varphi\le \frac{\pi}{2}\text{.}$ Then $\vr(\theta,\varphi)$ traces out one quarter of a circle starting at the north pole $\vr(\theta,0) = (0,0,1)$ (but excluding the north pole itself) and ending at the point $\vr(\theta,\frac{\pi}{2}) = (\cos\theta,\sin\theta,0)$ in the $xy$-plane.
Keep $\theta$ fixed at the same value and extend the interval over which $\varphi$ runs to $0 \lt \varphi \lt \pi\text{.}$ Now $\vr(\theta,\varphi)$ traces out a semi-circle starting at the north pole $\vr(\theta,0) = (0,0,1)\text{,}$ ending at the south pole $\vr(\theta,\pi) = (0,0,-1)$ (but excluding both the north and south poles themselves) and passing through the point $\vr(\theta,\frac{\pi}{2}) = (\cos\theta,\sin\theta,0)$ in the $xy$-plane.
Finally have $\theta$ run over $0\le\theta \lt 2\pi\text{.}$ Then the semicircle rotates about the $z$-axis, sweeping out the full sphere, except for the north and south poles.

Recall that $\varphi$ is the angle between the radius vector and the $z$-axis. If you hold $\varphi$ fixed and increase $\theta$ by a small amount $\dee{\theta}\text{,}$ $\vr(\theta,\varphi)$ sweeps out the red circular arc in the figure on the left below. If you hold $\varphi$ fixed and vary $\theta$ from $0$ to $2\pi\text{,}$ $\vr(\theta,\varphi)$ sweeps out a line of latitude. The figure on the right below gives the lines of latitude (or at least the parts of those lines in the first octant) for $\varphi=\frac{\pi}{10}\text{,}$ $\frac{2\pi}{10}\text{,}$ $\frac{3\pi}{10}\text{,}$ $\frac{4\pi}{10}$ and $\frac{5\pi}{10}=\frac{\pi}{2}\text{.}$

On the other hand, if you hold $\theta$ fixed and increase $\varphi$ by a small amount $\dee{\varphi}\text{,}$ $\vr(\theta,\varphi)$ sweeps out the red circular arc in the figure on the left below. If you hold $\theta$ fixed and vary $\varphi$ from $0$ to $\pi\text{,}$ $\vr(\theta,\varphi)$ sweeps out a line of longitude. The figure on the right below gives the lines of longitude (or at least the parts of those lines in the first octant) for $\theta=0\text{,}$ $\frac{\pi}{10}\text{,}$ $\frac{2\pi}{10}\text{,}$ $\frac{3\pi}{10}\text{,}$ $\frac{4\pi}{10}$ and $\frac{5\pi}{10}=\frac{\pi}{2}\text{.}$

Example 3.1.3. Cylinder.

The surface $x^2+z^2=1$ is an infinite cylinder. Part of this cylinder in the first octant is sketched below.

Note that the section of this cylinder that lies in the $xz$-plane, and in fact in any plane $y=c\text{,}$ is the circle $x^2+z^2=1\text{.}$ We can of course parametrize this circle by $x=\cos\theta\text{,}$ $z=\sin\theta\text{.}$ So we can parametrize the whole cylinder by using $\theta$ and $y$ as parameters.

\begin{equation*} \vr(\theta,y) = \big(\cos\theta\,,\,y\,,\,\sin\theta\big) \qquad 0\le\theta \lt 2\pi,\ \ -\infty \lt y \lt \infty \end{equation*}

Example 3.1.4. Surface of Revolution.

In this example, we are going to parametrize a surface of revolution. In your first integral calculus course, you undoubtedly encountered many surfaces created by rotating a curve $y=f(x)$ about the $x$-axis or the $y$-axis. In this course, we are used to having the $z$-axis, rather than the $y$-axis, run vertically.

So in this example, we’ll parametrize the surface constructed by rotating the curve

\begin{equation*} z=g(y)=e^y \qquad 0\le y\le 1 \end{equation*}

about the $z$-axis. Exactly the same procedure can be used to parametrize surfaces created by rotating about the $x$-axis or the $y$-axis too.

We start by just sketching the curve, considering the $yz$-plane as the plane $x=0$ in $\bbbr^3\text{.}$ The specified curve is the red curve in the figure below. Concentrate on any one point on that curve. It is the blue dot at $(0,Y,e^Y)$

in the figure. When our curve is rotated about the $z$-axis, the blue dot sweeps out a circle. The circle that the blue dot sweeps out

lies in the horizontal plane $z=e^Y$ and
is centred on the $z$-axis and
has radius $Y\text{.}$

We can parametrize the circle swept out in the usual way. Here is a top view of the circle, with the parameter, named $\theta\text{,}$ indicated.

The coordinates of the red dot are $\big(Y\sin\theta\,,\,Y\cos\theta\,,\,e^Y\big)\text{.}$ This also gives a parametrization of the surface of revolution

\begin{align*} x(Y,\theta) & = Y\sin\theta\\ y(Y,\theta) & = Y\cos\theta\\ z(Y,\theta) & = e^Y\\ &0\le Y\le 1,\qquad 0\le\theta \lt 2\pi \end{align*}

Notice, by way of checks, that

when $\theta=0\text{,}$
\begin{equation*} \big(x(Y,0)\,,\,y(Y,0)\,,\,z(Y,0)\big) =(0,Y,e^Y) \end{equation*}
runs over the entire desired curve (namely $z=g(y)\text{,}$ $0\le y\le 1$), when $Y$ runs over $0\le Y\le 1$ and
for any fixed $0\le Y\le 1\text{,}$ $\big(x(Y,\theta)\,,\,y(Y,\theta)\,,\,z(Y,\theta)\big)$ runs over the circle $x^2+y^2=Y^2\text{,}$ in the plane $z=e^Y\text{,}$ when $\theta$ runs over $0\le\theta \lt 2\pi\text{.}$

Also notice that

\begin{equation*} x(Y,\theta)^2 + y(Y,\theta)^2 = Y^2 \end{equation*}

so that

\begin{equation*} Y=\sqrt{x(Y,\theta)^2 + y(Y,\theta)^2} \end{equation*}

and

\begin{gather*} z(Y,\theta) =e^{Y} = e^{ \sqrt{x(Y,\theta)^2 + y(Y,\theta)^2} } \end{gather*}

That is, the surface of revolution is contained in the (infinite) surface

\begin{equation*} z=e^{\sqrt{x^2+y^2}} \end{equation*}

Remembering that $0\le Y\le 1\text{,}$ we have that $1\le z=e^Y \le e\text{.}$ Thus the surface of revolution is

\begin{equation*} z=e^{\sqrt{x^2+y^2}}\qquad 1\le z\le e \end{equation*}

Finally here is a sketch of the part of the surface in the first octant, $x,y,z\ge 0\text{.}$

Example 3.1.5. Torus.

In this example, we are going to parametrize a donut (well, its surface), or an inner tube.

The formal mathematical name for the surface of a donut is a torus. Our strategy will be to first parametrize the section of the torus in the right half of the $yz$-plane, and then built up the full torus by rotating the circle about the $z$-axis. The section is a circle, sketched below.

We’ll assume that the centre of the circle is a distance $R$ from the $z$-axis, and that the circle has radius $r\text{.}$ Then the red dot on the circle is at

\begin{align*} x&=0\\ y&= R + r\cos\theta\\ z&= r\sin\theta \end{align*}

In particular the red dot is a distance $r\sin\theta$ above the $xy$-plane and is a distance $R + r\cos\theta$ from the $z$-axis. So when we rotate the section about the $z$-axis, the red dot sweeps out a circle which is sketched below.

The circle that the red dot sweeps out

lies in the plane $z=r\sin\theta$ and
is centred on the $z$-axis and
has radius $\rho=R + r\cos\theta\text{.}$

We can parametrize the circle swept out in the usual way. Here is a top view of the circle, with the parameter, named $\psi\text{,}$ indicated.

So the parametrization of the circle swept out by the red dot, and also the parametrization of the torus, is

\begin{align*} x &= \rho\cos\psi = (R + r\cos\theta)\cos\psi\\ y &= \rho\sin\psi = (R + r\cos\theta)\sin\psi\\ z &= r\sin\theta \end{align*}

\begin{align*} &\vr(\theta,\psi) = (R + r\cos\theta)\cos\psi \ \hi + (R + r\cos\theta)\sin\psi \ \hj + r\sin\theta\ \hk\\ & 0\le\theta,\psi \lt 2\pi \end{align*}

Exercises Exercises

Exercise Group.

Exercises — Stage 1

1.

Parametrize the surface given by $z=e^{x+1}+xy$ in terms of $x$ and $y\text{.}$

2. (✳).

Let $S$ be the surface given by

\begin{equation*} \vr(u, v) = \big( u + v\,,\, u^2 + v^2 \,,\, u - v\big),\qquad -2 \le u \le 2,\ -2 \le v \le 2 \end{equation*}

This is a surface you are familiar with. What surface is it (it may be just a portion of one of the following)? sphere / helicoid / ellipsoid / saddle / parabolic bowl / cylinder / cone / plane

Exercise Group.

Exercises — Stage 2

3. (✳).

Suppose $S$ is the part of the hyperboloid $x^2 + y^2 - 2z^2 = 1$ that lies inside the cylinder $x^2 + y^2 = 9$ and above the plane $z = 1$ (i.e. for which $z \ge 1$).

Which of the following are parameterizations of $S\text{?}$

The vector function
\begin{equation*} \vr(u,v) = u\,\hi + v\,\hj +\frac{\sqrt{u^2+v^2-1}}{\sqrt{2}}\,\hk \end{equation*}
with domain $D=\Set{(u,v)}{ 2 \le u^2+v^2 \le 9}\text{.}$
The vector function
\begin{equation*} \vr(u,v) = u\sin v\,\hi - u\cos v\,\hj +\sqrt{\frac{u^2}{2}-\frac{1}{2}}\,\hk \end{equation*}
with domain $D=\Set{(u,v)}{ \sqrt{3} \le u\le 3,\ 0\le v\le 2\pi}\text{.}$
The vector function
\begin{equation*} \vr(u,v) = \sqrt{1+2v^2}\cos u\,\hi + \sqrt{1+2v^2} \sin u\,\hj +v\,\hk \end{equation*}
with domain $D=\Set{(u,v)}{ 0\le u\le 2\pi,\ 1\le v\le 2}\text{.}$
The vector function
\begin{equation*} \vr(u,v) = \sqrt{1+u}\sin v\,\hi + \sqrt{1+u} \cos v\,\hj +\sqrt{\frac{u}{2}}\,\hk \end{equation*}
with domain $D=\Set{(u,v)}{ 2\le u\le 8,\ 0\le v\le 2\pi}\text{.}$
The vector function
\begin{equation*} \vr(u,v) = \sqrt{u}\cos v\,\hi - \sqrt{u} \sin v\,\hj +\frac{\sqrt{u+1}}{\sqrt{2}}\,\hk \end{equation*}
with domain $D=\Set{(u,v)}{ 3\le u\le 9,\ 0\le v\le 2\pi}\text{.}$

4. (✳).

Suppose the surface $S$ is the part of the sphere $x^2 + y^2 + z^2 = 2$ that lies inside the cylinder $x^2 + y^2 = 1$ and for which $z \ge 0\text{.}$ Which of the following are parameterizations of $S\text{?}$

\begin{equation*} \vr(\phi,\theta) = 2\sin \phi\cos\theta\,\hi +2\cos \phi\,\hj +2\sin \phi\sin\theta\,\hk \end{equation*}

\begin{equation*} 0\le\phi\le\frac{\pi}{4},\ 0\le\theta\le2\pi \end{equation*}
\begin{equation*} \vr(x,y) = x\,\hi -y\,\hj +\sqrt{2-x^2-y^2}\,\hk \end{equation*}

\begin{equation*} x^2+y^2\le 1 \end{equation*}
\begin{equation*} \vr(u,\theta) = u\sin\theta\,\hi +u\cos \theta\,\hj +\sqrt{2-u^2}\,\hk \end{equation*}

\begin{equation*} 0\le u\le 2,\ 0\le\theta\le2\pi \end{equation*}
\begin{equation*} \vr(\phi,\theta) = \sqrt{2}\sin \phi\cos\theta\,\hi +\sqrt{2}\sin \phi\sin\theta\,\hj +\sqrt{2}\cos \phi\,\hk \end{equation*}

\begin{equation*} 0\le\phi\le\frac{\pi}{4},\ 0\le\theta\le2\pi \end{equation*}
\begin{equation*} \vr(\phi,z) = -\sqrt{2-z^2}\sin \phi\,\hi +\sqrt{2-z^2}\cos \phi\,\hj +z\,\hk \end{equation*}

\begin{equation*} 0\le\phi\le2\pi,\ 1\le z\le\sqrt{2} \end{equation*}

5. (✳).

Let $S$ be the part of the paraboloid $z + x^2 + y^2 = 4$ lying between the planes $z = 0$ and $z = 1\text{.}$ For each of the following, indicate whether or not it correctly parameterizes the surface $S\text{.}$

$\vr(u,v) = u\,\hi + v\,\hj + (4 - u^2 - v^2)\,\hk\text{,}$ $0 \le u^2 + v^2 \le 1$
$\vr(u,v) = (\sqrt{4-u}\,\cos v)\,\hi + (\sqrt{4-u}\, \sin v)\,\hj + u\,\hk\text{,}$ $0 \le u \le 1\text{,}$ $0 \le v \le 2\pi$
$\vr(u, v) = (u \cos v)\,\hi + (u \sin v)\,\hj + (4 - u^2 )\,\hk\text{,}$ $\sqrt{3} \le u \le 2\text{,}$ $0 \le v \le 2\pi$

Exercise Group.

Exercises — Stage 3

6. (✳).

Consider the following surfaces

$S_1$ is the hemisphere given by the equation $x^2 + y^2 + z^2 = 4$ with $z\ge 0\text{.}$
$S_2$ is the cylinder given by the equation $x^2 + y^2 = 1\text{.}$
$S_3$ is the cone given by the equation $z^2 = x^2 + y^2$ with $z\ge 0\text{.}$

Consider the following parameterizations:

\begin{align*} &\vr(\theta, \phi) =\big(\sqrt{4}\cos\theta\sin\phi\,,\, \sqrt{4}\sin\theta\sin\phi\,,\, \sqrt{4}\cos\phi\big)\\ & 0\le\theta\le2\pi,\quad 0\le\phi\le\pi/6 \end{align*}
\begin{align*} &\vr(\theta, \phi) =\big(\sqrt{4}\cos\theta\sin\phi\,,\, \sqrt{4}\sin\theta\sin\phi\,,\, \sqrt{4}\cos\phi\big)\\ & 0\le\theta\le2\pi,\quad 0\le\phi\le\pi/4 \end{align*}
\begin{align*} &\vr(\theta, \phi) =\big(\sqrt{4}\cos\theta\sin\phi\,,\, \sqrt{4}\sin\theta\sin\phi\,,\, \sqrt{4}\cos\phi\big)\\ & 0\le\theta\le2\pi,\quad 0\le\phi\le\pi/3 \end{align*}
\begin{align*} &\vr(\theta,z) = \big(\sqrt{4-z^2}\cos\theta\,,\,\sqrt{4-z^2}\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad 1\le z\le 2 \end{align*}
\begin{align*} &\vr(\theta,z) = \big(\sqrt{4-z^2}\cos\theta\,,\,\sqrt{4-z^2}\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad \sqrt{2}\le z\le 2 \end{align*}
\begin{align*} &\vr(\theta,z) = \big(\sqrt{4-z^2}\cos\theta\,,\,\sqrt{4-z^2}\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad \sqrt{3}\le z\le 2 \end{align*}
\begin{align*} &\vr(\theta,z) = \big(z\cos\theta\,,\,z\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad 0\le z\le 1 \end{align*}
\begin{align*} &\vr(\theta,z) = \big(z\cos\theta\,,\,z\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad 0\le z\le \sqrt{2} \end{align*}
\begin{align*} &\vr(\theta,z) = \big(z\cos\theta\,,\,z\sin\theta\,,\,z\big)\\ & 0\le\theta\le2\pi,\quad 0\le z\le \sqrt{3} \end{align*}
\begin{align*} &\vr(x,y) =\big(x\,,\,y\,,\,\sqrt{x^2+y^2}\big)\\ & x^2+y^2\le 1 \end{align*}
\begin{align*} &\vr(x,y) =\big(x\,,\,y\,,\,\sqrt{x^2+y^2}\big)\\ & x^2+y^2\le \sqrt{2} \end{align*}
\begin{align*} &\vr(x,y) =\big(x\,,\,y\,,\,\sqrt{x^2+y^2}\big)\\ & x^2+y^2\le 2 \end{align*}

For each of the following, choose from above all of the valid parameterization of each of the given surfaces. Note that there may be one or more valid parameterization for each surface, and not necessarily all of the above parameterizations will be used.

The part of $S_1$ contained inside $S_2\text{:}$
The part of $S_1$ contained inside $S_3\text{:}$
The part of $S_3$ contained inside $S_2\text{:}$
The part of $S_3$ contained inside $S_1\text{:}$

7.

Parametrize a solid of rotation about a line not parallel to an axis. Maybe first show that the plane you’re rotating is normal to that axis.

Give a parametric equation for the circle of radius 1, centred at $(2,2,4)\text{,}$ lying in the plane $x=y\text{.}$
Give a parametrized equation for the surface formed by rotating the circle from part (a) about the line $\vr(t)=4\hi+4\hj+t\hk\text{.}$

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