14.3 (optional) Sketching surfaces in 3D

Consider the traces. That is, if you set one variable equal to a constant, what will the resulting cross-sections look like?

(a) $\leftrightarrow$ (B) (b) $\leftrightarrow$ (A) (c) $\leftrightarrow$ (C)

(a) Each constant $z$ cross–section of $x^2+y^2=z^2+1$ is a (horizontal) circle centred on the $z$–axis. The radius of the circle is $1$ when $z=0$ and grows as $z$ moves away from $z=0$. So $x^2+y^2=z^2+1$ consists of a bunch of (horizontal) circles stacked on top of each other, with the radius increasing with $|z|$. It is a hyperboloid of one sheet. The picture that corresponds to (a) is (B).

(b) Every point of $y=x^2+z^2$ has $y\ge 0$. Only (A) has that property. We can also observe that every constant $y$ cross–section is a circle centred on $x=z=0$. The radius of the circle is zero when $y=0$ and increases as $y$ increases. The surface $y=x^2+z^2$ is a paraboloid. The picture that corresponds to (b) is (A).

(c) The only possibility left is that the picture that corresponds to (c) is (C).

Draw in the plane $z=C$ for several values of $C$.

We first add into the sketch of the graph the horizontal planes $z=C$, for $C=3$, $2$, $1$, $0.5$, $0.25$.

To reduce clutter, for each $C$, we have drawn in only

We have also omitted the label for the plane $z=0.25$.

The intersection of the plane $z=C$ with the graph $z=f(x,y)$ is line

Drawing this line (which is parallel to the $x$-axis) in the $xy$-plane, rather than in the plane $z=C$, gives a level curve. Doing this for each of $C=3$, $2$, $1$, $0.5$, $0.25$ gives five level curves.

Remember when you set $f(x,y)$ equal to a constant, the result is a curve with only $x$'s and $y$'s.

(a)

(b)

(c)

(a) For each fixed $c>0$, the level curve $x^2+2y^2=c$ is the ellipse centred on the origin with $x$ semi axis $\sqrt{c}$ and $y$ semi axis $\sqrt{c/2}$. If $c=0$, the level curve $x^2+2y^2=c=0$ is the single point $(0,0)$.

(b) For each fixed $c\ne 0$, the level curve $xy=c$ is a hyperbola centred on the origin with asymptotes the $x$- and $y$-axes. If $c>0$, any $x$ and $y$ obeying $xy=c>0$ are of the same sign. So the hyperbola is contained in the first and third quadrants. If $c<0$, any $x$ and $y$ obeying $xy=c>0$ are of opposite sign. So the hyperbola is contained in the second and fourth quadrants. If $c=0$, the level curve $xy=c=0$ is the single point $(0,0)$.

(c) For each fixed $c\ne 0$, the level curve $xe^{-y}=c$ is the logarithmic curve $y=-\ln\frac{c}{x}$. Note that, for $c>0$, the curve

and for $c<0$, the curve

If $c=0$, the level curve $xe^{-y}=c=0$ is the $y$-axis, $x=0$.

The circle centred at $(0,a)$ with radius $r$ has equation

Rearranged, this is

Use this to describe the level curves of the function given.

If $C=0$, the level curve $f=C=0$ is just the line $y=0$. If $C\ne 0$ (of either sign), we may rewrite the equation, $f(x,y)=\frac{2y}{x^2+y^2}=C$, of the level curve $f=C$ as

which is the equation of the circle of radius $\frac{1}{|C|}$ centred on $\left(0\,,\,\frac{1}{C}\right)$.

Remark.
To be picky, the function $f(x,y)=\frac{2y}{x^2+y^2}$ is not defined at $(x,y)=(0,0)$. The question should have either specified that the domain of $f$ excludes $(0,0)$ or have specified a value for $f(0,0)$. In fact, it is impossible to assign a value to $f(0,0)$ in such a way that $f(x,y)$ is continuous at $(0,0)$, because $\lim_{x\rightarrow 0}f(x,0)=0$ while $\lim_{y\rightarrow 0}f(0,|y|)=\infty$. So it makes more sense to have the domain of $f$ being $\bbbr^2$
with the point $(0,0)$ removed. That's why there is a little hole at the origin in the above sketch.

If $z$ is constant, then the entire expression $-z^2+2z$ is one big constant.

(a)

(b)

(a) We can rewrite the equation as

The right hand side is negative for $|z-1|<1$, i.e. for $0<z<2$. So no point on the surface has $0<z<2$. For any fixed $z$, outside that range, the curve $x^2 + y^2 = (z-1)^2 - 1$ is the circle of radius $\sqrt{(z-1)^2 - 1}$ centred on the $z$–axis. That radius is $0$ when $z=0,2$ and increases as $z$ moves away from $z=0,2$. For very large $|z|$, the radius increases roughly linearly with $|z|$. Here is a sketch of some level curves.

(b) The surface consists of two stacks of circles. One stack starts with radius $0$ at $z=2$. The radius increases as $z$ increases. The other stack starts with radius $0$ at $z=0$. The radius increases as $z$ decreases. This surface is a hyperboloid of two sheets. Here are two sketchs. The sketch on the left is of the part of the surface in the first octant. The sketch on the right of the full surface.

For each fixed $z$, $4x^2+y^2=1+z^2$ is an ellipse. So the surface consists of a stack of ellipses one on top of the other. The

For each fixed $z$, $4x^2+y^2=1+z^2$ is an ellipse. So the surface consists of a stack of ellipses one on top of the other. The semi axes are $\frac{1}{2}\sqrt{1+z^2}$ and $\sqrt{1+z^2}$. These are smallest when $z=0$ (i.e. for the ellipse in the $xy$-plane) and increase as $|z|$ increases. The intersection of the surface with the $xz$-plane (i.e. with the plane $y=0$) is the hyperbola $4x^2-z^2=1$ and the intersection with the $yz$-pane (i.e. with the plane $x=0$) is the hyperbola $y^2-z^2=1$. Here are two sketches of the surface. The sketch on the left only shows the part of the surface in the first octant (with axes).

Start by determining what convenient traces look like. For (a), the level curves are less instructive at first than are the traces found by setting $y$ equal to a constant.

(a)

(b)

(c)

(a) The graph is $z=\sin x$ with $(x,y)$ running over $0\le x\le 2\pi$, $0\le y\le 1$. For each fixed $y_0$ between $0$ and $1$, the intersection of this graph with the vertical plane $y=y_0$ is the same sin graph $z=\sin x$ with $x$ running from $0$ to $2\pi$. So the whole graph is just a bunch of 2-d sin graphs stacked side-by-side. This gives the graph on the left below.

(b) The graph is $z=\sqrt{x^2+y^2}$. For each fixed $z_0\ge 0$, the intersection of this graph with the horizontal plane $z=z_0$ is the circle $\sqrt{x^2+y^2}=z_0$. This circle is centred on the $z$-axis and has radius $z_0$. So the graph is the upper half of a cone. It is the sketch on the right above.

(c) The graph is $z=|x|+|y|$. For each fixed $z_0\ge 0$, the intersection of this graph with the horizontal plane $z=z_0$ is the square $|x|+|y|=z_0$. The side of the square with $x,y\ge 0$ is the straight line $x+y=z_0$. The side of the square with $x\ge 0$ and $y\le 0$ is the straight line $x-y=z_0$ and so on. The four corners of the square are $(\pm z_0,0,z_0)$ and $(0, \pm z_0,z_0)$. So the graph is a stack of squares. It is an upside down four-sided pyramid. The part of the pyramid in the first octant (that is, $x,y,z\ge 0$) is the sketch below.

(a) This is an elliptic cylinder parallel to the $z$-axis. Here is a sketch of the part of the surface above the $xy$–plane.

(b) This is a plane through $(4,0,0)$, $(0,4,0)$ and $(0,0,2)$. Here is a sketch of the part of the plane in the first octant.

(c) This is a hyperboloid of one sheet with axis the $x$-axis.

(d) This is a circular cone centred on the $y$-axis.

(e) This is an ellipsoid centered on the origin with semiaxes $3$, $\sqrt{12}=2\sqrt{3}$ and $3$ along the $x$, $y$ and $z$-axes, respectively.

(f) This is a sphere of radius $r_b=\frac{1}{2}\sqrt{b^2+4b+97}$ centered on $\frac{1}{2}(-4,b,-9)$.

(g) This is an elliptic paraboloid with axis the $x$-axis.

(h) This is an upward openning parabolic cylinder.

(a) For each fixed $z_0$, the $z=z_0$ cross-section (parallel to the $xy$-plane) of this surface is an ellipse centered on the origin with one semiaxis of length 2 along the $x$-axis and one semiaxis of length 4 along the $y$-axis. So this is an elliptic cylinder parallel to the $z$-axis. Here is a sketch of the part of the surface above the $xy$–plane.

(b) This is a plane through $(4,0,0)$, $(0,4,0)$ and $(0,0,2)$. Here is a sketch of the part of the plane in the first octant.

(c) For each fixed $x_0$, the $x=x_0$ cross-section parallel to the $yz$-plane is an ellipse with semiaxes $3\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $y$-axis and $2\sqrt{1+\frac{x_0^2}{16}}$ parallel to the $z$-axis. As you move out along the $x$-axis, away from $x=0$, the ellipses grow at a rate proportional to $\sqrt{1+\frac{x^2}{16}}$, which for large $x$ is approximately $\frac{|x|}{4}$. This is called a hyperboloid of one sheet. Its

(d) For each fixed $y_0$, the $y=x_0$ cross-section (parallel to the $xz$-plane) is a circle of radius $|y|$ centred on the $y$-axis. When $y_0=0$ the radius is $0$. As you move further from the $xz$-plane, in either direction, i.e. as $|y_0|$ increases, the radius grows linearly. The full surface consists of a bunch of these circles stacked sideways. This is a circular cone centred on the $y$-axis.

(e) This is an ellipsoid centered on the origin with semiaxes $3$, $\sqrt{12}=2\sqrt{3}$ and $3$ along the $x$, $y$ and $z$-axes, respectively.

(f) Completing three squares, we have that $x^2+y^2+z^2+4x-by+9z-b=0$ if and only if $(x+2)^2+\big(y-\frac{b}{2}\big)^2+\big(z+\frac{9}{2}\big)^2 =b+4+\frac{b^2}{4}+\frac{81}{4}$. This is a sphere of radius $r_b=\frac{1}{2}\sqrt{b^2+4b+97}$ centered on $\frac{1}{2}(-4,b,-9)$.

(g) There are no points on the surface with $x<0$. For each fixed $x_0>0$ the cross-section $x=x_0$ parallel to the $yz$-plane is an ellipse centred on the $x$–axis with semiaxes $\sqrt{x_0}$ in the $y$-axis direction and $\frac{3}{2}\sqrt{x_0}$ in the $z$–axis direction. As you increase $x_0$, i.e. move out along the $x$-axis, the ellipses grow at a rate proportional to $\sqrt{x_0}$. This is an elliptic paraboloid with axis the $x$-axis.

(h) This is called a parabolic cylinder. For any fixed $y_0$, the
$y=y_0$ cross-section (parallel to the $xz$-plane) is the upward opening parabola $z=x^2$ which has vertex on the $y$-axis.

To solve (say) $\sin(x+y)=0$, you get lots of solutions: $x+y=0$, $x+y=\pi$, $x+y=2\pi$, etc.

$z=0$:

$z=1$:

$z=2$:

The level curves of $z=0$ correspond to all points $(x,y)$ such that $0=\sin(x+y)$. The angles that make $\sin \theta$ equal to 0 are $\theta = \pi n$ for integer values of $n$. So, the level curves are lines of the form

where $n$ is any integer.

So, our level curve has the lines $y=-x$, $y=\pi-x$, $y=2\pi-x$, etc.

The level curves of $z=1$ correspond to all points $(x,y)$ such that $1=\sin(x+y)$. The angles that make $\sin \theta$ equal to 1 are $\theta =\frac{pi}{2}+ 2\pi n$ for integer values of $n$. So, the level curves are lines of the form

where $n$ is any integer.

So, our level curve has the lines $y=\frac{\pi}{2}-x$, $y=\frac{\pi}{2}+2\pi-x$, $y=\frac{\pi}{2}+4\pi-x$, etc.

The equation $2=\sin(x+y)$ has no solutions, since no angle has sine greater than 1. So the level curve at $z=2$ has no points:

Since the level curves are circles centred at the origin (in the $xy$-plane), the equation will have the form $x^2+y^2=g(z)$, where $g(z)$ is a function depending only on $z$.

$\displaystyle x^2+y^2=\left( \frac{|z|}{3}+1\right)^2$

Since the level curves are circles centred at the origin (in the $xy$-plane), when $z$ is a constant, the equation will have the form $x^2+y^2=c$ for some constant. That is, our equation looks like

where $g(z)$ is a function depending only on $z$.

Because our cross-sections are so nicely symmetric, we know the intersection of the figure with the left side of the $yz$-plane as well: $z=3(-y-1)=-3(y+1)$ (when $z\ge0$) and $z=-3(-y-1)=3(y+1)$ (when $z<0$). Below is the intersection of our surface with the $yz$ plane.

Setting $x=0$, our equation becomes $y^2=g(z)$. Looking at the right side of the $yz$ plane, this should lead to: $\left.\begin{cases} z=3(y-1) &\text{if } z\geq 0,\ y\ge 1\ z=-3(y-1) &\text{if } z< 0,\ y\ge 1 \end{cases}\right}$. That is:

A quick check: when we squared both sides of the equation in $(*)$, we added another solution, $\frac{|z|}{3}+1=-y$. Let's make sure we haven't diverged from our diagram.

This matches our diagram eactly. So, all together, the equation of the surface is