Sketching Surfaces in 3d

Section 1.7 Sketching Surfaces in 3d

In practice students taking multivariable calculus regularly have great difficulty visualising surfaces in three dimensions, despite the fact that we all live in three dimensions. We’ll now develop some technique to help us sketch surfaces in three dimensions

Of course you could instead use some fancy graphing software, but part of the point is to build intuition. Not to mention that you can’t use fancy graphing software on your exam.

We all have a fair bit of experience drawing curves in two dimensions. Typically the intersection of a surface (in three dimensions) with a plane is a curve lying in the (two dimensional) plane. Such an intersection is usually called a cross-section. In the special case that the plane is one of the coordinate planes, the intersection is sometimes called a trace. One can often get a pretty good idea of what a surface looks like by sketching a bunch of cross-sections. Here are some examples.

Example 1.7.1. \(4x^2+y^2-z^2=1\).

Sketch the surface that satisfies \(4x^2+y^2-z^2=1\text{.}\)

Solution.

We’ll start by fixing any number \(z_0\) and sketching the part of the surface that lies in the horizontal plane \(z=z_0\text{.}\)

The intersection of our surface with that horizontal plane is a horizontal cross-section. Any point \((x,y,z)\) lying on that horizontal cross-section satisfies both

\begin{align*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2-z^2=1\\ \iff &z=z_0\ \ \text{and}\ \ 4x^2+y^2=1+z_0^2 \end{align*}

Think of \(z_0\) as a constant. Then \(4x^2+y^2=1+z_0^2\) is a curve in the \(xy\)-plane. As \(1+z_0^2\) is a constant, the curve is an ellipse. To determine its semi-axes

The semi-axes of an ellipse are the line segments from the centre of the ellipse to the farthest points on the ellipse and to the nearest points on the ellipse. For a circle the lengths of all of these line segments are just the radius.

, we observe that when \(y=0\text{,}\) we have \(x=\pm\frac{1}{2}\sqrt{1+z_0^2}\) and when \(x=0\text{,}\) we have \(y=\pm\sqrt{1+z_0^2}\text{.}\) So the curve is just an ellipse with \(x\) semi-axis \(\frac{1}{2}\sqrt{1+z_0^2}\) and \(y\) semi-axis \(\sqrt{1+z_0^2}\text{.}\) It’s easy to sketch.

Remember that this ellipse is the part of our surface that lies in the plane \(z=z_0\text{.}\) Imagine that the sketch of the ellipse is on a single sheet of paper. Lift the sheet of paper up, move it around so that the \(x\)- and \(y\)-axes point in the directions of the three dimensional \(x\)- and \(y\)-axes and place the sheet of paper into the three dimensional sketch at height \(z_0\text{.}\) This gives a single horizontal ellipse in 3d, as in the figure below.

We can build up the full surface by stacking many of these horizontal ellipses — one for each possible height \(z_0\text{.}\) So we now draw a few of them as in the figure below. To reduce the amount of clutter in the sketch, we have only drawn the first octant (i.e. the part of three dimensions that has \(x\ge 0\text{,}\) \(y\ge 0\) and \(z\ge 0\)).

Here is why it is OK, in this case, to just sketch the first octant. Replacing \(x\) by \(-x\) in the equation \(4x^2+y^2-z^2=1\) does not change the equation. That means that a point \((x,y,z)\) is on the surface if and only if the point \((-x,y,z)\) is on the surface. So the surface is invariant under reflection in the \(yz\)-plane. Similarly, the equation \(4x^2+y^2-z^2=1\) does not change when \(y\) is replaced by \(-y\) or \(z\) is replaced by \(-z\text{.}\) Our surface is also invariant under reflection in the \(xz\)- and \(xy\)-planes. Once we have the part in the first octant, the remaining octants can be gotten simply by reflecting about the coordinate planes.

We can get a more visually meaningful sketch by adding in some vertical cross-sections. The \(x=0\) and \(y=0\) cross-sections (also called traces — they are the parts of our surface that are in the \(yz\)- and \(xz\)-planes, respectively) are

\begin{equation*} x=0,\ y^2-z^2=1\qquad\text{and}\qquad y=0,\ 4x^2-z^2=1 \end{equation*}

These equations describe hyperbolae

It’s not just a figure of speech!

. If you don’t remember how to sketch them, don’t worry. We’ll do it now. We’ll first sketch them in 2d. Since

\begin{alignat*}{2} y^2&=1+z^2 & \quad\implies\quad &|y|\ge 1\\ &&&\text{ and }\quad y=\pm 1\text{ when }z=0\\ &&&\text{ and }\quad\text{for large } z,\ y\approx\pm z\\ 4x^2&=1+z^2 & \quad\implies\quad &|x|\ge \tfrac{1}{2}\\ &&&\text{ and }\quad x=\pm\tfrac{1}{2}\text{ when }z=0\\ &&&\text{ and }\quad\text{for large } z,\ x\approx\pm \tfrac{1}{2}z \end{alignat*}

the sketches are

Now we’ll incorporate them into the 3d sketch. Once again imagine that each is a single sheet of paper. Pick each up and move it into the 3d sketch, carefully matching up the axes. The red (blue) parts of the hyperbolas above become the red (blue) parts of the 3d sketch below (assuming of course that you are looking at this on a colour screen).

Now that we have a pretty good idea of what the surface looks like we can clean up and simplify the sketch. Here are a couple of possibilities.

Here are two figures created by graphing software.

This type of surface is called a hyperboloid of one sheet.

There are also hyperboloids of two sheets. For example, replacing the \(+1\) on the right hand side of \(x^2+y^2-z^2=1\) gives \(x^2+y^2-z^2=-1\text{,}\) which is a hyperboloid of two sheets. We’ll sketch it quickly in the next example.

Example 1.7.2. \(4x^2+y^2-z^2=-1\).

Sketch the surface that satisfies \(4x^2+y^2-z^2=-1\text{.}\)

Solution.

As in the last example, we’ll start by fixing any number \(z_0\) and sketching the part of the surface that lies in the horizontal plane \(z=z_0\text{.}\) The intersection of our surface with that horizontal plane is

\begin{align*} &z=z_0\ \ \text{and}\ \ 4x^2+y^2=z_0^2-1 \end{align*}

Think of \(z_0\) as a constant.

If \(|z_0| \lt 1\text{,}\) then \(z_0^2-1 \lt 0\) and there are no solutions to \(x^2+y^2=z_0^2-1\text{.}\)
If \(|z_0|=1\) there is exactly one solution, namely \(x=y=0\text{.}\)
If \(|z_0| \gt 1\) then \(4x^2+y^2=z_0^2-1\) is an ellipse with \(x\) semi-axis \(\frac{1}{2}\sqrt{z_0^2-1}\) and \(y\) semi-axis \(\sqrt{z_0^2-1}\text{.}\) These semi-axes are small when \(|z_0|\) is close to \(1\) and grow as \(|z_0|\) increases.

The first octant parts of a few of these horizontal cross-sections are drawn in the figure below.

Next we add in the \(x=0\) and \(y=0\) cross-sections (i.e. the parts of our surface that are in the \(yz\)- and \(xz\)-planes, respectively)

\begin{equation*} x=0,\ z^2=1+y^2\qquad\text{and}\qquad y=0,\ z^2=1+4x^2 \end{equation*}

Now that we have a pretty good idea of what the surface looks like we clean up and simplify the sketch.

Here is are two figures created by graphing software.

This type of surface is called a hyperboloid of two sheets.

Example 1.7.3. \(yz=1\).

Sketch the surface \(yz=1\text{.}\)

Solution.

This surface has a special property that makes it relatively easy to sketch. There are no \(x\)’s in the equation \(yz=1\text{.}\) That means that if some \(y_0\) and \(z_0\) obey \(y_0z_0=1\text{,}\) then the point \((x,y_0,z_0)\) lies on the surface \(yz=1\) for all values of \(x\text{.}\) As \(x\) runs from \(-\infty\) to \(\infty\text{,}\) the point \((x,y_0,z_0)\) sweeps out a straight line parallel to the \(x\)-axis. So the surface \(yz=1\) is a union of lines parallel to the \(x\)-axis. It is invariant under translations parallel to the \(x\)-axis. To sketch \(yz=1\text{,}\) we just need to sketch its intersection with the \(yz\)-plane and then translate the resulting curve parallel to the \(x\)-axis to sweep out the surface.

We’ll start with a sketch of the hyperbola \(yz=1\) in two dimensions.

Next we’ll move this 2d sketch into the \(yz\)-plane, i.e. the plane \(x=0\text{,}\) in 3d, except that we’ll only draw in the part in the first octant.

The we’ll draw in \(x=x_0\) cross-sections for a couple of more values of \(x_0\)

and clean up the sketch a bit

Here are two figures created by graphing software.

Example 1.7.4. \(xyz=4\).

Sketch the surface \(xyz=4\text{.}\)

Solution.

We’ll sketch this surface using much the same procedure as we used in Examples 1.7.1 and 1.7.2. We’ll only sketch the part of the surface in the first octant. The remaining parts (in the octants with \(x,y \lt 0\text{,}\) \(z\ge 0\text{,}\) with \(x,z \lt 0\text{,}\) \(y\ge 0\) and with \(y,z \lt 0\text{,}\) \(x\ge0\)) are just reflections of the first octant part.

As usual, we start by fixing any number \(z_0\) and sketching the part of the surface that lies in the horizontal plane \(z=z_0\text{.}\) The intersection of our surface with that horizontal plane is the hyperbola

\begin{align*} &z=z_0\ \ \text{and}\ \ xy=\frac{4}{z_0} \end{align*}

Note that \(x\rightarrow\infty\) as \(y\rightarrow 0\) and that \(y\rightarrow\infty\) as \(x\rightarrow 0\text{.}\) So the hyperbola has both the \(x\)-axis and the \(y\)-axis as asymptotes, when drawn in the \(xy\)-plane. The first octant parts of a few of these horizontal cross-sections (namely, \(z_0=4\text{,}\) \(z_0=2\) and \(z_0=\frac{1}{2}\)) are drawn in the figure below.

Next we add some vertical cross-sections. We can’t use \(x=0\) or \(y=0\) because any point on \(xyz=4\) must have all of \(x\text{,}\) \(y\text{,}\) \(z\) nonzero. So we use

\begin{equation*} x=4,\ yz=1\qquad\text{and}\qquad y=4,\ xz=1 \end{equation*}

instead. They are again hyperbolae.

Finally, we clean up and simplify the sketch.

Here are two figures created by graphing software.

Subsection 1.7.1 Level Curves and Surfaces

Often the reason you are interested in a surface in 3d is that it is the graph \(z=f(x,y)\) of a function of two variables \(f(x,y)\text{.}\) Another good way to visualize the behaviour of a function \(f(x,y)\) is to sketch what are called its level curves. By definition, a level curve of \(f(x,y)\) is a curve whose equation is \(f(x,y)=C\text{,}\) for some constant \(C\text{.}\) It is the set of points in the \(xy\)-plane where \(f\) takes the value \(C\text{.}\) Because it is a curve in 2d, it is usually easier to sketch than the graph of \(f\text{.}\) Here are a couple of examples.

Example 1.7.5. \(f(x,y) = x^2+4y^2-2x+2\).

Sketch the level curves of \(f(x,y) = x^2+4y^2-2x+2\text{.}\)

Solution.

Fix any real number \(C\text{.}\) Then, for the specified function \(f\text{,}\) the level curve \(f(x,y)=C\) is the set of points \((x,y)\) that obey

\begin{align*} x^2+4y^2-2x+2=C &\iff x^2-2x+1 + 4y^2 +1 =C\\ &\iff (x-1)^2 + 4y^2 = C-1 \end{align*}

Now \((x-1)^2 + 4y^2\) is the sum of two squares, and so is always at least zero. So if \(C-1 \lt 0\text{,}\) i.e. if \(C \lt 1\text{,}\) there is no curve \(f(x,y)=C\text{.}\) If \(C-1=0\text{,}\) i.e. if \(C=1\text{,}\) then \(f(x,y)=C-1=0\) if and only if both \((x-1)^2=0\) and \(4y^2=0\) and so the level curve consists of the single point \((1,0)\text{.}\) If \(C \gt 1\text{,}\) then \(f(x,y)=C\) become \((x-1)^2+4y^2=C-1 \gt 0\) which describes an ellipse centred on \((1,0)\text{.}\) It intersects the \(x\)-axis when \(y=0\) and

\begin{equation*} (x-1)^2 = C-1 \iff x-1=\pm\sqrt{C-1} \iff x=1\pm \sqrt{C-1} \end{equation*}

and it intersects the line \(x=1\) (i.e. the vertical line through the centre) when

\begin{equation*} 4y^2 = C-1 \iff 2y=\pm\sqrt{C-1} \iff y=\pm\tfrac{1}{2} \sqrt{C-1} \end{equation*}

So, when \(C \gt 1\text{,}\) \(f(x,y)=C\) is the ellipse centred on \((1,0)\) with \(x\) semi-axis \(\sqrt{C-1} \) and \(y\) semi-axis \(\tfrac{1}{2}\sqrt{C-1}\text{.}\) Here is a sketch of some representative level curves of \(f(x,y) = x^2+4y^2-2x+2\text{.}\)

It is often easier to develop an understanding of the behaviour of a function \(f(x,y)\) by looking at a sketch of its level curves, than it is by looking at a sketch of its graph. On the other hand, you can also use a sketch of the level curves of \(f(x,y)\) as the first step in building a sketch of the graph \(z=f(x,y)\text{.}\) The next step would be to redraw, for each \(C\text{,}\) the level curve \(f(x,y)=C\text{,}\) in the plane \(z=C\text{,}\) as we did in Example 1.7.1.

Example 1.7.6. \(e^{x+y+z}=1\).

The function \(f(x,y)\) is given implicitly by the equation \(e^{x+y+z}=1\text{.}\) Sketch the level curves of \(f\text{.}\)

Solution.

This one is not as nasty as it appears. That “\(f(x,y)\) is given implicitly by the equation \(e^{x+y+z}=1\)” means that, for each \(x,y\text{,}\) the solution \(z\) of \(e^{x+y+z}=1\) is \(f(x,y)\text{.}\) So, for the specified function \(f\) and any fixed real number \(C\text{,}\) the level curve \(f(x,y)=C\) is the set of points \((x,y)\) that obey

\begin{align*} e^{x+y+C}=1 &\iff x+y+C = 0\qquad\text{(by taking the logarithm of both sides)}\\ &\iff x+y = -C \end{align*}

This is of course a straight line. It intersects the \(x\)-axis when \(y=0\) and \(x=-C\) and it intersects the \(y\)-axis when \(x=0\) and \(y=-C\text{.}\) Here is a sketch of some level curves.

We have just seen that sketching the level curves of a function \(f(x,y)\) can help us understand the behaviour of \(f\text{.}\) We can generalise this to functions \(F(x,y,z)\) of three variables. A level surface of \(F(x,y,z)\) is a surface whose equation is of the form \(F(x,y,z)=C\) for some constant \(C\text{.}\) It is the set of points \((x,y,z)\) at which \(F\) takes the value \(C\text{.}\)

Example 1.7.7. \(F(x,y,z)=x^2+y^2+z^2\).

Let \(F(x,y,z)=x^2+y^2+z^2\text{.}\) If \(C \gt 0\text{,}\) then the level surface \(F(x,y,z)=C\) is the sphere of radius \(\sqrt{C}\) centred on the origin. Here is a sketch of the parts of the level surfaces \(F=1\) (radius \(1\)), \(F=4\) (radius \(2\)) and \(F=9\) (radius \(3\)) that are in the first octant.

Example 1.7.8. \(F(x,y,z)=x^2+z^2\).

Let \(F(x,y,z)=x^2+z^2\) and \(C \gt 0\text{.}\) Consider the level surface \(x^2+z^2=C\text{.}\) The variable \(y\) does not appear in this equation. So for any fixed \(y_0\text{,}\) the intersection of our surface \(x^2+z^2=C\) with the plane \(y=y_0\) is the circle of radius \(\sqrt{C}\) centred on \(x=z=0\text{.}\) Here is a sketch of the first quadrant part of one such circle.

The full surface is the horizontal stack of all of those circles with \(y_0\) running over \(\bbbr\text{.}\) It is the cylinder of radius \(\sqrt{C}\) centred on the \(y\)-axis. Here is a sketch of the parts of the level surfaces \(F=1\) (radius \(1\)), \(F=4\) (radius \(2\)) and \(F=9\) (radius \(3\)) that are in the first octant.

Example 1.7.9. \(F(x,y,z)=e^{x+y+z}\).

Let \(F(x,y,z)=e^{x+y+z}\) and \(C \gt 0\text{.}\) Consider the level surface \(e^{x+y+z}=C\text{,}\) or equivalently, \(x+y+z=\ln C\text{.}\) It is the plane that contains the intercepts \((\ln C,0,0)\text{,}\) \((0,\ln C,0)\) and \((0,0,\ln C)\text{.}\) Here is a sketch of the parts of the level surfaces

\(F=e\) (intercepts \((1,0,0)\text{,}\) \((0,1,0)\text{,}\) \((0,0,1)\)),
\(F=e^2\) (intercepts \((2,0,0)\text{,}\) \((0,2,0)\text{,}\) \((0,0,2)\)) and
\(F=e^3\) (intercepts \((3,0,0)\text{,}\) \((0,3,0)\text{,}\) \((0,0,3)\))

that are in the first octant.

Exercises 1.7.2 Exercises

Exercise Group.

Exercises — Stage 1

1. (✳).

Match the following equations and expressions with the corresponding pictures. Cartesian coordinates are \((x, y, z)\text{,}\) cylindrical coordinates are \((r, \theta, z)\text{,}\) and spherical coordinates are \((\rho, \theta, \varphi)\text{.}\)

(A)

(B)

(C)

(D)

(E)

(F)

\begin{alignat*}{7} &\text{(a)}\quad& \varphi&=\pi/3 & &\text{(b)}\quad& r&=2\cos\theta & &\text{(c)}\quad& x^2+y^2&=z^2+1\\ &\text{(d)}& y&=x^2+z^2\qquad & &\text{(e)}& \rho&=2\cos\varphi\qquad & &\text{(f)}& z&=x^4+y^4-4xy & \end{alignat*}

2.

In each of (a) and (b) below, you are provided with a sketch of the first quadrant parts of a few level curves of some function \(f(x,y)\text{.}\) Sketch the first octant part of the corresponding graph \(z=f(x,y)\text{.}\)

(a)

(b)

3.

Sketch a few level curves for the function \(f(x,y)\) whose graph \(z=f(x,y)\) is sketched below.

Exercise Group.

Exercises — Stage 2

4.

Sketch some of the level curves of

\(\displaystyle f(x,y)=x^2+2y^2\)
\(\displaystyle f(x,y)=xy\)
\(\displaystyle f(x,y)=xe^{-y}\)

5. (✳).

Sketch the level curves of \(f(x,y)=\frac{2y}{x^2+y^2}\text{.}\)

6. (✳).

Draw a “contour map” of \(f(x, y) = e^{-x^2 +4y^2}\) , showing all types of level curves that occur.

7. (✳).

A surface is given implicitly by

\begin{equation*} x^2 + y^2 - z^2 + 2z = 0 \end{equation*}

Sketch several level curves \(z = \)constant.
Draw a rough sketch of the surface.

8. (✳).

Sketch the hyperboloid \(z^2=4x^2+y^2-1\text{.}\)

9.

Describe the level surfaces of

\(\displaystyle f(x,y,z)=x^2+y^2+z^2\)
\(\displaystyle f(x,y,z)=x+2y+3z\)
\(\displaystyle f(x,y,z)=x^2+y^2\)

10.

Sketch the graphs of

\(\displaystyle f(x,y)=\sin x\qquad 0\le x\le 2\pi,\ 0\le y\le 1\)
\(\displaystyle f(x,y)=\sqrt{x^2+y^2}\)
\(\displaystyle f(x,y)=|x|+|y|\)

11.

Sketch and describe the following surfaces.

\(\displaystyle 4x^2+y^2=16\)
\(\displaystyle x+y+2z=4\)
\(\displaystyle \frac{y^2}{9}+\frac{z^2}{4}=1+\frac{x^2}{16}\)
\(\displaystyle y^2=x^2+z^2\)
\(\displaystyle \frac{x^2}{9}+\frac{y^2}{12}+\frac{z^2}{9}=1\)
\(x^2+y^2+z^2+4x-by+9z-b=0\) where \(b\) is a constant.
\(\displaystyle \frac{x}{4}=\frac{y^2}{4}+\frac{z^2}{9}\)
\(\displaystyle z=x^2\)

Exercise Group.

Exercises — Stage 3

12.

The surface below has circular level curves, centred along the \(z\)-axis. The lines given are the intersection of the surface with the right half of the \(yz\)-plane. Give an equation for the surface.

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