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a. $\frac{{x}^{2}}{{3963}^{2}}+\frac{{y}^{2}}{{3963}^{2}}+\frac{{z}^{2}}{{3950}^{2}}=1;$

b.

;

c. The intersection curve is the ellipse of equation $\frac{{x}^{2}}{{3963}^{2}}+\frac{{y}^{2}}{{3963}^{2}}=\frac{(2950)(4950)}{{3950}^{2}},$ and the intersection is an ellipse.; d. The intersection curve is the ellipse of equation $\frac{2{y}^{2}}{{3963}^{2}}+\frac{{z}^{2}}{{3950}^{2}}=1.$

a.

b. The intersection curve is ${({x}^{2}+{z}^{2}-1)}^{3}-{x}^{2}{z}^{3}=0.$

(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.

(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

• is restricted to $x>0$, so that $\frac{c}{x}>0$ and $\ln \frac{c}{x}$ is defined, and that
• as $x\rightarrow 0^+$, $y$ goes to $-\infty$, while
• as $x\rightarrow +\infty$, $y$ goes to $+\infty$, and
• the curve crosses the $x$-axis (i.e. has $y=0$) when $x=c$.

and for $c<0$, the curve

• is restricted to $x<0$, so that $\frac{c}{x}>0$ and $\ln \frac{c}{x}$ is defined, and that
• as $x\rightarrow 0^-$, $y$ goes to $-\infty$, while
• as $x\rightarrow -\infty$, $y$ goes to $+\infty$, and
• the curve crosses the $x$-axis (i.e. has $y=0$) when $x=c$.

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

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

$$x^2-\frac{2}{C}y+y^2=0 \iff x^2+\left(y-\frac{1}{C}\right)^2 =\frac{1}{C^2}$$

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 $\mathbb{R}^2$ with the point $(0,0)$ removed. That's why there is a little hole at the origin in the above sketch.

(a) We can rewrite the equation as

$$x^2 + y^2 = (z-1)^2 - 1$$

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 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).

(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.