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(a) $\varphi =\frac{\pi}{3}$ is a surface of constant (spherical coordinate) $\varphi$. So it is a cone with vertex at the origin. We can express $\varphi=\frac{\pi}{3}$ in cartesian coordinates by observing that $0\le\varphi\le\frac{\pi}{2}$ so that $z\ge 0$, and

$$\begin{aligned} \varphi=\frac{\pi}{3} \iff \tan\varphi =\frac{\sqrt{3}}{2} \iff \rho\sin\varphi =\frac{\sqrt{3}}{2}\rho\cos\varphi \iff \sqrt{x^2+y^2}=\frac{\sqrt{3}}{2} z \end{aligned}$$

So the picture that corresponds to (a) is (C).

(b) As $r$ and $\theta$ are cylindrical coordinates

$$\begin{aligned} r=2\cos\theta \iff r^2=2 r\cos\theta \iff x^2+y^2 = 2x \iff (x-1)^2+y^2=1 \end{aligned}$$

There is no $z$ appearing in $(x-1)^2+y^2=1$. So every constant $z$ cross–section of $(x-1)^2+y^2=1$ is a (horizontal) circle of radius $1$ centred on the line $x=1$, $y=0$. It is a cylinder of radius $1$ centred on the line $x=1$, $y=0$. So the picture that corresponds to (b) is (F).

(c) 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 (c) is (D).

(d) Every point of $y=x^2+z^2$ has $y\ge 0$. Only (B) 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 (d) is (B).

(e) As $\rho$ and $\varphi$ are spherical coordinates

$$\begin{aligned} \rho=2\cos\varphi \iff \rho^2=2\rho\cos\varphi \iff x^2+y^2+z^2=2z \iff x^2+y^2 + (z-1)^2=1 \end{aligned}$$

This is the sphere of radius $1$ centred on $(0,0,1)$. The picture that corresponds to (e) is (A).

(f) The only possibility left is that the picture that corresponds to (f) is (E).

Each solution below consists of three sketchs.

• In the first sketch, we just redraw the given level curves with the $x$- and $y$-axes reoriented so that the sketch looks like we are high on the $z$-axis looking down at the $xy$-plane.
• In the second sketch, we lift up each level curve $f(x,y)=C$ and draw it in the horizontal plane $z=C$. That is we draw

$$\begin{aligned} &\big\{ \ (x,y,z)\ \big|\ f(x,y)=C,\ z=C,\ x\ge 0, y\ge 0\ \big\} \\ &=\big\{ \ (x,y,z)\ \big|\ z=f(x,y),\ z=C,\ x\ge 0, y\ge 0\ \big\} \end{aligned}$$

• Finally, in the third sketch, we draw the part of graph $z=f(x,y)$ in the first octant, just by “filling in the gaps in the second sketch”.

(a)

(b)

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

• the (gray) intersection of the horizontal plane $z=C$ with the $yz$–plane, i.e. with the vertical plane $x=0$, and
• the (blue) intersection of the horizontal plane $z=C$ with the graph $z=f(x,y)$.

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

$$\big\{ \ (x,y,z)\ \big|\ z=f(x,y),\ z=C\ \big\} = \big\{ \ (x,y,z)\ \big|\ f(x,y)=C,\ z=C\ \big\}$$

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.