Solution.

(a)

In \(x=k\): \(z = k^2+y^2\), an upward parabola in the \(yz-\)plane, raised by \(k^2\). In \(y=k\): \(z = x^2+k^2\), likewise. In \(z=k\): \(x^2+y^2 = k\), a circle of radius \(\sqrt k\) for \(k\geq 0\), a single point for \(k=0\), and nothing for \(k<0\).

Stacking circles of radius \(\sqrt k\) at height \(k\), with parabolic vertical sections, gives a circular paraboloid opening upward with vertex at the origin.

(b)

\(g = -f\), so the graph is the reflection of \(f\) in the plane \(z=0\): the same paraboloid opening downward.

(c)

\(h = g+3\), so the graph is that of \(g\) translated three units upward: a downward paraboloid with vertex \((0,0,3)\).

Solution.

(a)

divide by \(144\): \[\frac {x^2}{9} - \frac {y^2}{16} + \frac {z^2}{4} = 1 .\] Two positive terms and one negative, with the right-hand side positive: a hyperboloid of one sheet, with axis along \(y\), the variable carrying the minus sign. The trace in \(y=k\) is the ellipse \(\dfrac {x^2}{9}+\dfrac {z^2}{4} = 1+\dfrac {k^2}{16}\), which exists for every \(k\) and grows with \(\left |k\right |\) — so the surface is connected, as one sheet requires.

(b)

Only one variable appears to the first power, so this is a paraboloid opening along \(x\). The trace in \(x=k\) is the ellipse \(y^2+4z^2 = k\), present only for \(k\geq 0\): an elliptic paraboloid with vertex at the origin, opening in the positive \(x\) direction.

Solution.

(a)

In \(z=k\): \(x^2+y^2 = 1+k^2\), circles that widen as \(\left |k\right |\) grows. In \(x=k\): \(y^2-z^2 = 1-k^2\), hyperbolas, degenerating to the pair of lines \(y=\pm z\) when \(k=\pm 1\). In \(y=k\) likewise. The surface is a hyperboloid of one sheet about the \(z-\)axis.

(b)

The same surface with the roles of \(y\) and \(z\) exchanged: a hyperboloid of one sheet about the \(y-\)axis. The shape is unchanged; only the axis moves, since it is always the variable carrying the minus sign.

(c)

Completing the square in \(y\), \[x^2+(y+1)^2-z^2 = 1 ,\] the same hyperboloid of one sheet, translated one unit in the negative \(y\) direction. Its axis is the line \(x=0\), \(y=-1\).

Solution. The paraboloid \(z = x^2+y^2\) meets the cylinder \(x^2+y^2 = 1\) where \(z=1\), in the circle of radius \(1\) at that height.

The region is bounded below by the paraboloid, on the outside by the cylinder, and above by the plane \(z=2\). In cylindrical coordinates it is \[\big \{(r,\theta ,z): 0\leq r\leq 1,\ 0\leq \theta \leq 2\pi ,\ r^2\leq z\leq 2\big \} \quad \text {together with}\quad 1\leq z\leq 2 ,\] that is, the solid between the bowl of the paraboloid and the plane \(z=2\), contained within the cylinder — a cylindrical tub with a paraboloidal floor.

Solution. The two meet where \(x^2+y^2 = 2-x^2-y^2\), that is \(x^2+y^2 = 1\), at height \(z = 1\).

The first opens upward from the origin, the second downward from \((0,0,2)\). The region between them is a lens-shaped solid, circular in horizontal section, with its widest circle of radius \(1\) at height \(1\), closing to points at \((0,0,0)\) below and \((0,0,2)\) above.

Solution. Complete the square in each variable, watching the sign on \(y\): \[\big (x^2-4x\big ) - \big (y^2+2y\big ) + \big (z^2-2z\big ) + 4 = 0\] \[(x-2)^2-4 - \Big [(y+1)^2-1\Big ] + (z-1)^2-1 + 4 = 0\] \[(x-2)^2 - (y+1)^2 + (z-1)^2 = 0 .\] The right-hand side is zero, not a positive constant, so this is not a hyperboloid but a cone, with axis parallel to the \(y-\)axis and vertex — the point usually called the centre — at \((2,-1,1)\).

The degenerate case is easy to miss: the same equation with \(1\) on the right would be a hyperboloid of one sheet, and with \(-1\) a hyperboloid of two sheets. The cone is the surface separating those two families.

Solution. Revolving about an axis replaces the distance from that axis by the square root of the sum of the squares of the two coordinates perpendicular to it.

(a)

About \(z\), the coordinate \(x\) is replaced by \(\sqrt {x^2+y^2}\): \[x^2+y^2+4z^2 = 16 ,\] an ellipsoid of revolution (oblate, since the \(z\) semi-axis is \(2\) against \(4\)).

(b)

About \(z\), \(y\) is replaced by \(\sqrt {x^2+y^2}\): \[z = 4-\big (x^2+y^2\big ),\] a downward circular paraboloid with vertex \((0,0,4)\).

(c)

About \(x\), \(z\) is replaced by \(\sqrt {y^2+z^2}\): \[y^2+z^2-x^2 = 1 ,\] a hyperboloid of one sheet about the \(x-\)axis.

(d)

About \(y\), \(z\) is replaced by \(\sqrt {x^2+z^2}\): \[\sqrt {x^2+z^2} = e^{-y^2},\qquad \text {that is}\qquad x^2+z^2 = e^{-2y^2},\] a bell-shaped surface of revolution, widest at \(y=0\) where the radius is \(1\).

Solution.

(a)

With \(\rho = \sqrt 8 = 2\sqrt 2\), \(\phi = \dfrac {\pi }{3}\), \(\theta = \dfrac {\pi }{4}\), \[x = \rho \sin \phi \cos \theta = 2\sqrt 2\cdot \frac {\sqrt 3}{2}\cdot \frac {1}{\sqrt 2} = \sqrt 3 ,\] and by the same calculation with \(\sin \theta \), \(y = \sqrt 3\), while \[z = \rho \cos \phi = 2\sqrt 2\cdot \frac 12 = \sqrt 2 .\] So the rectangular coordinates are \(\big (\sqrt 3, \sqrt 3, \sqrt 2\big )\).

In cylindrical coordinates \(r = \sqrt {x^2+y^2} = \sqrt 6\), the angle is unchanged at \(\theta = \dfrac {\pi }{4}\), and \(z = \sqrt 2\), giving \(\Big (\sqrt 6, \dfrac {\pi }{4}, \sqrt 2\Big )\).

(b)

The sphere gives \(\rho = \sqrt {32} = 4\sqrt 2\). The plane \(z=4\) gives \(\rho \cos \phi = 4\), so \[\cos \phi = \frac {4}{4\sqrt 2} = \frac {1}{\sqrt 2}\implies \phi = \frac {\pi }{4}.\] In spherical coordinates the circle is therefore \(\rho = 4\sqrt 2\), \(\phi = \dfrac {\pi }{4}\), with \(\theta \) free.

In cylindrical coordinates, \(r^2 = 32-z^2 = 32-16 = 16\), so it is \(r = 4\), \(z = 4\), again with \(\theta \) free.

Solution. Multiply both sides by \(\rho \), which introduces no new points other than the origin: \[\rho ^2\sin ^2\phi = \rho \cos \phi .\] Now \(\rho \sin \phi \) is the distance from the \(z-\)axis, so \(\rho ^2\sin ^2\phi = x^2+y^2\), and \(\rho \cos \phi = z\). Hence \[x^2+y^2 = z ,\] a circular paraboloid opening upward with vertex at the origin.

Recognising \(\rho \sin \phi \) and \(\rho \cos \phi \) as the cylindrical radius and the height is what makes the conversion immediate.

Solution. The cone \(z = \sqrt {x^2+y^2}\) makes an angle of \(\dfrac {\pi }{4}\) with the positive \(z-\)axis, since there \(\rho \cos \phi = \rho \sin \phi \).

”Below the cone” and ”above the \(xy-\)plane” therefore means \(\phi \) runs from \(\dfrac {\pi }{4}\) to \(\dfrac {\pi }{2}\), and the sphere bounds \(\rho \). So \[E = \Big \{(\rho ,\phi ,\theta ): 0\leq \rho \leq \sqrt 7,\ \frac {\pi }{4}\leq \phi \leq \frac {\pi }{2},\ 0\leq \theta \leq 2\pi \Big \},\] the part of the ball lying outside the cone and above the equatorial plane — a shape like a thick lens with a conical bite taken out of its upper face.

Solution. The point satisfies the equation: \(16-2-2 = 12\). With \(F = z^2-2x^2-2y^2-12\), \[\nabla F = \langle -4x, -4y, 2z\rangle \ \longrightarrow \ \langle -4, 4, 8\rangle .\] The tangent plane through \((1,-1,4)\) with this normal is \[-4(x-1) + 4(y+1) + 8(z-4) = 0 ,\] which simplifies, on dividing by \(4\), to \[-x + y + 2z = 6 .\]

Solution. The point satisfies the equation: \(2(1)(4) - 3(1)(-1) - 4 = 8+3-4 = 7\). With \(G = 2xz^2-3xy-4x-7\), \[\nabla G = \langle 2z^2-3y-4,\ -3x,\ 4xz\rangle \ \longrightarrow \ \langle 8+3-4,\ -3,\ 8\rangle = \langle 7,-3,8\rangle .\] Hence the tangent plane is \[7(x-1) - 3(y+1) + 8(z-2) = 0 ,\] that is \(7x-3y+8z = 26\).

Solution.

(a)

Completing the square in each variable, \[9(x-2)^2 - 36 + 4(y+1)^2 - 4 + 18(z+3)^2 - 162 - 158 = 0 ,\] \[9(x-2)^2 + 4(y+1)^2 + 18(z+3)^2 = 360 ,\] and dividing by \(360\), \[\frac {(x-2)^2}{40} + \frac {(y+1)^2}{90} + \frac {(z+3)^2}{20} = 1 .\] All three terms are positive, so this is an ellipsoid, with centre \((2,-1,-3)\) and semi-axes \(\sqrt {40}\), \(\sqrt {90}\), \(\sqrt {20}\).

(b)

The point given does not lie on the surface. Substituting \(x=3\), \(y=2\), \(z = 3+\dfrac {1}{\sqrt 2}\) into the left-hand side gives \[9(1) + 4(9) + 18\Big (6+\tfrac {1}{\sqrt 2}\Big )^2 \approx 854.7 ,\] against the required \(360\); equivalently the original expression evaluates to \(\big (18+3\sqrt 2\big )^2\approx 494.7\) rather than \(0\). Since the centre has \(z=-3\) and the \(z\) semi-axis is only \(\sqrt {20}\approx 4.47\), no point of the surface has \(z\) anywhere near \(3.7\).

The tangent plane is therefore computed at the point of the surface with the same \(x\) and \(y\). Putting \(x=3\), \(y=2\) gives \(18(z+3)^2 = 360-45 = 315\), so \(z = -3+\dfrac {\sqrt {70}}{2}\approx 1.183\). There \[\nabla F = \langle 18x-36,\ 8y+8,\ 36z+108\rangle = \big \langle 18,\ 24,\ 18\sqrt {70}\big \rangle ,\] and dividing by \(6\), a normal is \(\big \langle 3, 4, 3\sqrt {70}\big \rangle \). The tangent plane is \[3(x-3) + 4(y-2) + 3\sqrt {70}\Big (z+3-\frac {\sqrt {70}}{2}\Big ) = 0 ,\] that is \[3x + 4y + 3\sqrt {70}\,z = 122 - 9\sqrt {70} .\]

Solution. The point satisfies the equation, since \(2\cdot 3\cdot 1 = 6\). With \(F = xyz-6\), \[\nabla F = \langle yz, xz, xy\rangle \ \longrightarrow \ \langle 3, 2, 6\rangle .\] The normal line runs through \((2,3,1)\) in that direction, so \[\frac {x-2}{3} = \frac {y-3}{2} = \frac {z-1}{6}.\]

Solution. The point satisfies the equation: \(4(-2)+2(2)(3) = -8+12 = 4\). With \(K = x^2y+2xz-4\), \[\nabla K = \langle 2xy+2z,\ x^2,\ 2x\rangle \ \longrightarrow \ \langle -8+6,\ 4,\ 4\rangle = \langle -2, 4, 4\rangle .\] Its length is \(\sqrt {4+16+16} = 6\), so the unit normals are \[\pm \frac {1}{6}\langle -2,4,4\rangle = \pm \Big \langle -\frac 13, \frac 23, \frac 23\Big \rangle .\]

Solution. \[\nabla T = \langle -8x, -2y\rangle \ \longrightarrow \ \nabla T(2,-3) = \langle -16, 6\rangle .\] The particle should move along the gradient, in the direction of the unit vector \[\frac {1}{\sqrt {256+36}}\langle -16, 6\rangle = \frac {1}{2\sqrt {73}}\langle -16, 6\rangle = \frac {1}{\sqrt {73}}\langle -8, 3\rangle ,\] and the rate of increase in that direction is \(\left |\nabla T\right | = 2\sqrt {73}\approx 17.1\) degrees per unit length.

The direction points back towards the origin, where the plate is hottest, which is the check worth making.

Solution. Both pass through the point: \(4+1+4 = 9\), and \(4+1-3 = 2\).

The angle between two surfaces at a common point is the angle between their normals there. For the sphere, \(\nabla (x^2+y^2+z^2) = \langle 2x,2y,2z\rangle \) gives \(\textbf {n}_1 = \langle 4,-2,4\rangle \). For the paraboloid, written as \(x^2+y^2-z-3 = 0\), the gradient is \(\langle 2x,2y,-1\rangle \), giving \(\textbf {n}_2 = \langle 4,-2,-1\rangle \). Then \[\cos \theta = \frac {16+4-4}{\sqrt {36}\sqrt {21}} = \frac {16}{6\sqrt {21}} = \frac {8\sqrt {21}}{63}\approx 0.582 ,\] so \(\theta \approx 54.4^{\circ }\).