Solution. Along the curve, \(\varphi \) becomes a function of \(t\) alone through \(x(t)\), \(y(t)\), \(z(t)\). The chain rule for a function of three variables gives \[\frac {d\varphi }{dt} = \frac {\partial \varphi }{\partial x}\frac {dx}{dt} + \frac {\partial \varphi }{\partial y}\frac {dy}{dt} + \frac {\partial \varphi }{\partial z}\frac {dz}{dt}.\] The right-hand side is precisely the dot product of \[\nabla \varphi = \Big \langle \frac {\partial \varphi }{\partial x}, \frac {\partial \varphi }{\partial y}, \frac {\partial \varphi }{\partial z}\Big \rangle \qquad \text {with}\qquad \frac {d\overline {r}}{dt} = \Big \langle \frac {dx}{dt}, \frac {dy}{dt}, \frac {dz}{dt}\Big \rangle ,\] which is the stated identity. It is the reason the directional derivative is a dot product with the gradient: a direction is just the velocity of a curve through the point.

Solution.

(a)

\[\frac {\partial M}{\partial y} = 12x^3y^2 = \frac {\partial N}{\partial x},\] so the field is conservative. Integrating \(M\) with respect to \(x\), \[f = x^4y^3 + \ln \left |x\right | + g(y).\] Differentiating with respect to \(y\) and matching \(N\), \[3x^4y^2 + g'(y) = 3x^4y^2 - \frac 1y \implies g(y) = -\ln \left |y\right | .\] Hence \[f = x^4y^3 + \ln \left |x\right | - \ln \left |y\right | + C .\]

(b)

Here the test is that the curl vanishes: \[\nabla \times \textbf {F} = \big \langle xe^z-xe^z,\ ye^z-ye^z,\ e^z-e^z\big \rangle = \textbf {0}.\] Integrating the first component with respect to \(x\) gives \(f = xye^z + g(y,z)\), and the other two components then force \(g\) to be constant. Hence \[f = xye^z + C .\]

Solution.

(a)

Writing \(\textbf {u} = \langle u_1,u_2,u_3\rangle \) and applying the product rule to each term, \[\nabla \cdot (\Phi \textbf {u}) = \sum _{i}\frac {\partial }{\partial x_i}\big (\Phi u_i\big ) = \sum _i\Big (\frac {\partial \Phi }{\partial x_i}u_i + \Phi \frac {\partial u_i}{\partial x_i}\Big ) = \nabla \Phi \cdot \textbf {u} + \Phi \big (\nabla \cdot \textbf {u}\big ).\] It is the ordinary product rule, applied three times and collected.

(b)

Since \(r = \sqrt {x^2+y^2+z^2}\) we have \(\dfrac {\partial r}{\partial x} = \dfrac {x}{r}\), so \[\nabla r^n = nr^{n-1}\Big \langle \frac xr,\frac yr,\frac zr\Big \rangle = nr^{n-2}\,\overline {r}.\] Now apply part (a) with \(\Phi = nr^{n-2}\) and \(\textbf {u} = \overline {r}\), noting \(\nabla \cdot \overline {r} = 3\): \[\nabla \cdot \big (nr^{n-2}\overline {r}\big ) = n(n-2)r^{n-4}\big (\overline {r}\cdot \overline {r}\big ) + 3nr^{n-2} = n(n-2)r^{n-2} + 3nr^{n-2},\] so \[\dive \grad r^n = n(n+1)r^{n-2}.\] Note that \(n = -1\) gives zero: that is the statement that \(\dfrac 1r\) is harmonic, which is why it is the potential of a point source.

Solution.

(a)

\[\nabla \cdot \textbf {F} = 2xyz + 2xyz + 2xyz = 6xyz ,\] \[\nabla \times \textbf {F} = \big \langle x\big (z^2-y^2\big ),\ y\big (x^2-z^2\big ),\ z\big (y^2-x^2\big ) \big \rangle .\]

(b)

Here the \(\textbf {j}\) component is zero. \[\nabla \cdot \textbf {F} = ye^z + 0 + ye^x = y\big (e^x+e^z\big ),\] \[\nabla \times \textbf {F} = \big \langle ze^x,\ xye^z - yze^x,\ -xe^z\big \rangle .\]

(c)

This is \(\dfrac {\overline {r}}{r}\), the unit radial field. Using part (a) of the previous question with \(\Phi = r^{-1}\), \[\nabla \cdot \textbf {F} = \frac 2r = \frac {2}{\sqrt {x^2+y^2+z^2}},\] and the curl is \(\textbf {0}\), since the field is the gradient of \(r\).

(d)

\[\nabla \cdot \textbf {F} = \frac 1y + \frac 1z + \frac 1x ,\] \[\nabla \times \textbf {F} = \Big \langle \frac {y}{z^2},\ \frac {z}{x^2},\ \frac {x}{y^2}\Big \rangle .\]

Solution. \[\nabla f = \big \langle 2xy,\ x^2+z^2,\ 2zy\big \rangle .\]

(a)

\[\nabla \cdot (\nabla f) = 2y + 0 + 2y = 4y ,\] which is the Laplacian \(f_{xx}+f_{yy}+f_{zz}\).

(b)

\[\nabla \times (\nabla f) = \big \langle 2z-2z,\ 0-0,\ 2x-2x\big \rangle = \textbf {0}.\] This is no accident: the curl of any gradient is zero, because mixed partial derivatives are equal. It is the reason a field with non-zero curl can never be conservative.

Solution. First test for path independence: \[\frac {\partial M}{\partial y} = 1 = \frac {\partial N}{\partial x},\] so the field is conservative and all four answers must agree. A potential is found by integrating: \[f = \frac {x^2}{2} + xy - \frac {y^2}{2}.\] Indeed \(f_x = x+y\) and \(f_y = x-y\). Hence, for every path, \[\int _{(1,1)}^{(4,2)} F\cdot d\overline {r} = f(4,2) - f(1,1) = \Big (8+8-2\Big ) - \Big (\tfrac 12+1-\tfrac 12\Big ) = 14 - 1 = 13 .\] The point of setting four paths is exactly that the labour of parametrising each one is unnecessary once the field is recognised as conservative.

Solution. By Green’s theorem the integral equals \[\iint _D\Big (\frac {\partial N}{\partial x}-\frac {\partial M}{\partial y}\Big )dA = \iint _D\big (3-(-1)\big )dA = 4\,\text {Area}(D).\] The triangle has base \(3\) and height \(2\), so its area is \(3\) and the integral is \[4\times 3 = 12 .\] Doing it directly requires three parametrisations and considerably more arithmetic for the same answer.

Solution.

(a)

\[\overline {r}\,'(t) = \langle 1, -2\sin 2t, 2\cos 2t\rangle ,\qquad \left |\overline {r}\,'\right | = \sqrt {1+4} = \sqrt 5 ,\] a constant. On the curve \(y^2+z^2 = 1\), so the integrand is \(t^2+1\) and \[\int _C = \sqrt 5\int _0^{2\pi }\big (t^2+1\big )dt = \sqrt 5\Big [\frac {t^3}{3}+t\Big ]_0^{2\pi } = \frac {2\sqrt 5\,\pi \big (4\pi ^2+3\big )}{3}\approx 199 .\]

(b)

The field is conservative: it is \(\nabla f\) with \[f = xy+yz+zx ,\] as differentiating confirms. So the integral depends only on the endpoints, and the intermediate point is irrelevant: \[f(0,1,2) - f(0,0,0) = \big (0+2+0\big ) - 0 = 2 .\]

Solution. \[\frac {\partial M}{\partial y} = 2x-4y^3 = \frac {\partial N}{\partial x},\] so on any simply connected region the field is conservative and the integral is path independent. Integrating \(M\) with respect to \(x\), \[f = x^2y - xy^4 + 3x + g(y),\] and matching \(f_y = x^2-4xy^3\) forces \(g\) constant. Hence \[\int _{(1,0)}^{(2,1)} F\cdot d\overline {r} = f(2,1)-f(1,0) = \big (4-2+6\big ) - \big (0-0+3\big ) = 8-3 = 5 .\]

Solution. Integrating the first component with respect to \(x\) gives \(f = xyz + g(y,z)\). Then \[f_y = xz + g_y = xz \implies g_y = 0 ,\] so \(g\) depends only on \(z\), and \[f_z = xy + g'(z) = xy+2z \implies g'(z) = 2z \implies g = z^2 .\] Hence \[f = xyz + z^2 + C .\]

Solution.

The double integral

\[\frac {\partial N}{\partial x} - \frac {\partial M}{\partial y} = -2y - \big (-3xy^2\big ) = 3xy^2 - 2y ,\] so \[\iint _D\big (3xy^2-2y\big )dA = \int _0^2\int _0^2\big (3xy^2-2y\big )dx\,dy = \int _0^2\big (6y^2-4y\big )dy = 16-8 = 8 .\]

The line integral

Traversing the square anticlockwise:

On \(y=0\), \(x: 0\rightarrow 2\), \(dy=0\): \(\int _0^2x^2dx = \dfrac 83\).

On \(x=2\), \(y: 0\rightarrow 2\), \(dx=0\): \(\int _0^2\big (y^2-4y\big )dy = \dfrac 83-8 = -\dfrac {16}{3}\).

On \(y=2\), \(x: 2\rightarrow 0\), \(dy=0\): \(\int _2^0\big (x^2-8x\big )dx = -\dfrac 83+16 = \dfrac {40}{3}\).

On \(x=0\), \(y: 2\rightarrow 0\), \(dx=0\): \(\int _2^0y^2dy = -\dfrac 83\).

Adding, \(\dfrac 83 - \dfrac {16}{3} + \dfrac {40}{3} - \dfrac 83 = \dfrac {24}{3} = 8\), which agrees with the double integral, as Green’s theorem requires.

Solution. First locate the parameters. From \(u^2+1 = 5\) and \(v^3+1 = 2\) we get \(u = \pm 2\) and \(v = 1\); the third coordinate \(u+v = 3\) selects \(u = 2\).

\[\overline {r}_u = \langle 2u, 0, 1\rangle = \langle 4,0,1\rangle ,\qquad \overline {r}_v = \big \langle 0, 3v^2, 1\big \rangle = \langle 0,3,1\rangle ,\] \[\overline {r}_u\times \overline {r}_v = \langle 0\cdot 1 - 1\cdot 3,\ 1\cdot 0 - 4\cdot 1,\ 12-0\rangle = \langle -3, -4, 12\rangle .\] The tangent plane through \((5,2,3)\) is therefore \[-3(x-5) - 4(y-2) + 12(z-3) = 0 ,\] that is \(-3x-4y+12z = 13\), or \(3x+4y-12z = -13\).

Solution.

(a)

\[\overline {r}_u = \langle 1,-3,1\rangle ,\qquad \overline {r}_v = \langle 1,0,-1\rangle ,\] \[\overline {r}_u\times \overline {r}_v = \langle 3,2,3\rangle ,\qquad \left |\overline {r}_u\times \overline {r}_v\right | = \sqrt {9+4+9} = \sqrt {22}.\] The cross product is constant, as it must be for a plane, so the area is simply \[S = \sqrt {22}\times \text {area of the parameter square} = 4\sqrt {22}\approx 18.8 .\]

(b)

Treating \(y\) as a function of \(x\) and \(z\), \(y_x = 4\) and \(y_z = 2z\), so \[S = \int _0^1\int _0^1\sqrt {1+16+4z^2}\,dx\,dz = \int _0^1\sqrt {17+4z^2}\,dz .\] With \(2z = \sqrt {17}\sinh \theta \) this evaluates to \[S = \frac {\sqrt {21}}{2} + \frac {17}{4}\sinh ^{-1}\frac {2}{\sqrt {17}} \approx 4.28 .\]

Solution. \[\overline {r}_u = \langle 1,1,2\rangle ,\qquad \overline {r}_v = \langle 1,-1,1\rangle ,\] \[\overline {r}_u\times \overline {r}_v = \langle 3,1,-2\rangle ,\qquad \left |\overline {r}_u\times \overline {r}_v\right | = \sqrt {14},\] again constant. On the surface, \[f = (u+v)+(u-v)+(1+2u+v) = 4u+v+1 ,\] so \[\iint _S f\,dS = \sqrt {14}\int _0^1\int _0^2\big (4u+v+1\big )du\,dv = \sqrt {14}\int _0^1\big (8+2v+2\big )dv = 11\sqrt {14}\approx 41.2 .\]

Solution.

The curl side

\[\nabla \times \textbf {F} = \langle 0,0,2\rangle .\] The cone opens upward and its boundary is the circle \(x^2+y^2 = 16\) at \(z=4\). Rather than integrate over the cone itself, note that Stokes’ theorem lets the surface be replaced by any other with the same boundary; the flat disc \(x^2+y^2\leq 16\) at \(z=4\) is simplest, with upward unit normal \(\textbf {k}\): \[\iint _S\big (\nabla \times \textbf {F}\big )\cdot d\textbf {S} = \iint _D 2\,dA = 2\pi (4)^2 = 32\pi .\]

The line integral

Parametrise the boundary anticlockwise as \(x = 4\cos t\), \(y = 4\sin t\), \(z = 4\), \(0\leq t\leq 2\pi \). Then \[\oint _C\textbf {F}\cdot d\overline {r} = \int _0^{2\pi }\Big [(-4\sin t)(-4\sin t) + (4\cos t)(4\cos t) + (-2)(0)\Big ]dt\] \[= \int _0^{2\pi }16\big (\sin ^2t+\cos ^2t\big )dt = 32\pi .\] The two agree, verifying the theorem.

Solution. \[\nabla \times \textbf {F} = \big \langle 3-2,\ 0-0,\ 0-x\big \rangle = \langle 1, 0, -x\rangle .\] Take \(S\) to be the part of the plane \(z = 5-x\) inside the cylinder. Writing it as \(g(x,y) = 5-x\), the upward normal element is \[d\textbf {S} = \langle -g_x, -g_y, 1\rangle dA = \langle 1, 0, 1\rangle dA .\] Hence \[\big (\nabla \times \textbf {F}\big )\cdot d\textbf {S} = \big (1 + 0 - x\big )dA = (1-x)\,dA ,\] and integrating over the disc \(x^2+y^2\leq 9\), \[\int _C\textbf {F}\cdot d\overline {r} = \iint _D(1-x)\,dA = \text {Area} - \iint _D x\,dA = 9\pi - 0 = 9\pi ,\] the second integral vanishing because \(x\) is odd over a disc centred at the origin.

Solution. Write \(r = \left |\overline {r}\right |\), so \(\textbf {F} = r\,\overline {r}\). By the product rule for divergence, \[\nabla \cdot \big (r\,\overline {r}\big ) = \nabla r\cdot \overline {r} + r\big (\nabla \cdot \overline {r}\big ) = \frac {\overline {r}\cdot \overline {r}}{r} + 3r = r + 3r = 4r ,\] using \(\nabla r = \dfrac {\overline {r}}{r}\) and \(\nabla \cdot \overline {r} = 3\). Therefore \[\iint _S\textbf {F}\cdot d\textbf {S} = \iiint _E 4r\,dV = \int _0^{2\pi }\int _0^{\pi }\int _0^a 4\rho \cdot \rho ^2\sin \phi \,d\rho \,d\phi \,d\theta = 4\pi a^4 .\] As a check, the field has constant magnitude \(a^2\) on the sphere and points radially outward, so the flux is \(a^2\times 4\pi a^2 = 4\pi a^4\) directly.

Solution. By symmetry \(\bar x = \bar y = 0\), so only \(\bar z\) is needed. In spherical coordinates the solid hemisphere is \(0\leq \rho \leq a\), \(0\leq \phi \leq \dfrac {\pi }{2}\), \(0\leq \theta \leq 2\pi \), and \(z = \rho \cos \phi \). The volume is \(\dfrac 23\pi a^3\), and \[\iiint z\,dV = \int _0^{2\pi }\int _0^{\pi /2}\int _0^a \rho \cos \phi \cdot \rho ^2\sin \phi \,d\rho \,d\phi \,d\theta = 2\pi \cdot \frac {a^4}{4}\cdot \frac 12 = \frac {\pi a^4}{4}.\] Hence \[\bar z = \frac {\pi a^4/4}{\frac 23\pi a^3} = \frac {3a}{8},\] and the centre of mass is \(\Big (0,0,\dfrac {3a}{8}\Big )\).

Solution. For the cone \(z = \sqrt {x^2+y^2}\), \[z_x = \frac {x}{\sqrt {x^2+y^2}},\qquad z_y = \frac {y}{\sqrt {x^2+y^2}},\] so \(1+z_x^2+z_y^2 = 2\) and the surface element is \(dS = \sqrt 2\,dA\) — constant, which is what makes cones convenient.

In polar coordinates \(z = r\), and \(1\leq z\leq 4\) becomes \(1\leq r\leq 4\), so \[m = \int _0^{2\pi }\int _1^4\big (10-r\big )\sqrt 2\;r\,dr\,d\theta = 2\pi \sqrt 2\int _1^4\big (10r-r^2\big )dr .\] Evaluating, \[\int _1^4\big (10r-r^2\big )dr = \Big [5r^2-\frac {r^3}{3}\Big ]_1^4 = \Big (80-\frac {64}{3}\Big ) - \Big (5-\frac 13\Big ) = 54 ,\] so \[m = 108\sqrt 2\,\pi \approx 480 .\]