Math 214 Follow-Up Assignment

May 13, 2002

1. Find the gradient vector of the function f that is defined by

f(x, y, z) = x^{4}y^{2} − y^{3}z^{2} + zx .

Response:

(∇f)(x, y, z) = (4x^{3}y^{2} + z, 2x^{4}y − 3y^{2}z^{2}, −2y^{3}z + x) .

2. Find the divergence of the vector field F that is defined by

F (x, y, z) = (z^{3}x, x^{3}y, y^{3}z) .

Response:

\frac{∂}{∂x}(z^{3}x) + \frac{∂}{∂y} (x^{3}y) + \frac{∂}{∂z} (y^{3}z) = z^{3} + x^{3} + y^{3} .

3. Find the double integral of the function f(x, y) = xy^{2} on the rectangle, having sides parallel to the coordinate axes, with diagonally opposite vertices at the points (−1, 1) and (5, 3).

Response: The rectangle, say E, is described by the inequalities

\{ \begin{matrix}−1 & ≤ & x & ≤ & 5 \\ 1 & ≤ & y & ≤ & 3 \end{matrix} .

Thus,

∫ ∫_{E} xy^{2}dA = ∫_{−1}^{5} ∫_{1}^{3} xy^{2}dydx

= ∫_{−1}^{5} x (∫_{1}^{3} y^{2}dy) dx

= ∫_{−1}^{5} x · [\frac{y^{3}}{3} ]_{y=1}^{y = 3} dx

= ∫_{−1}^{5} x · (9 − \frac{1}{3}) dx

= \frac{26}{3} ∫_{−1}^{5} xdx

= \frac{26}{3} [\frac{x^{2}}{2} ]_{x=−1}^{x = 5}

= \frac{26}{3} (\frac{5^{2}}{2} − \frac{(−1)^{2}}{2} )

= 104 .

4. Find the tangent plane at the point (3, −1, 2) to the surface

xy^{2}z = 6 .

Response: The chief task here is to find a vector normal to the tangent plane to be used as coefficient vector in the equation for the plane. Since the surface is a level set of the function

f(x, y, z) = xy^{2}z ,

the gradient ∇f of f at the given point must be normal to the tangent plane at that point (unless it vanishes).

(∇f)(x, y, z) = (y^{2}z, 2xyz, xy^{2})
(∇f)(3, −1, 2) = (2, −12, 3)

Hence, the tangent plane has an equation of the form

2x − 12y + 3z = constant .

Using the fact that (3, −1, 2) must satisfy this equation, one obtains the equation

2x − 12y + 3z = 24 .

5. Find the arc length of the helix that is given parametrically by

\{ \begin{matrix}x = 4 sin t \\ y = 4 cos t \\ z = 3t \end{matrix}

for 0 ≤ t ≤ π/2.

Response: This involves a straightforward application of the definition of the length of a parameterized path as the integral of the length of the derivative of a point on the path with respect to the the parameter:

∫_{C} ds = ∫_{C} ||(dx, dy, dz)||

= ∫_{0}^{π/2} ||(\frac{dx}{dt}, \frac{dy}{dt} , \frac{dz}{dt} )|| dt

= ∫_{0}^{π/2} ||(4 cos t, −4 sin t, 3)|| dt

= ∫_{0}^{π/2} \sqrt{4^{2} (cos^{2} t + sin^{2} t) + 3^{2}}dt

= ∫_{0}^{π/2} 5dt

= \frac{5π}{2} .

6. Find the path integral of the vector field F that is defined by

F (x, y, z) = (x − 2z, y + z, 1 − y)

over the path given by

R(t) = (t^{3}, t, t^{2})

for 0 ≤ t ≤ 2.

Response: This involves a straightforward application of the definition of the integral over a parameterized path of a vector field:

∫_{C} F dR = ∫_{0}^{2} F (R(t)) · R′(t)dt

= ∫_{0}^{2} (t^{3} − 2t^{2}, t + t^{2}, 1 − t) · (3t^{2}, 1, 2t) dt

= ∫_{0}^{2} ((3t^{5} − 6t^{4}) + (t^{2} + t) + (−2t^{2} + 2t)) dt

= ∫_{0}^{2}(3t^{5} − 6t^{4} − t^{2} + 3t) dt

= [\frac{t^{6}}{2} − \frac{6t^{5}}{5} − \frac{t^{3}}{3} + \frac{3t^{2}}{2} ]_{t=0}^{t = 2}

= 32 − \frac{192}{5} − \frac{8}{3} + 6

= −\frac{46}{15} .

7. Find the equation of the plane containing the point (−1, 3, 2) that is normal to the line defined by the two equations

\{ \begin{matrix}3x − 4y + 2z & = & 7 \\ 9x − 5y − 3z & = & 2 \end{matrix} .

Response: A key fact for this problem is that a single linear equation in 3 variables is the equation of a plane in space, and the coefficient vector of such an equation is a (unique up to a scalar multiple) normal vector to that plane.

Since a point in the required plane is given, one only needs to find a vector normal to the plane, which is the same thing as a vector parallel to the given line. The line is given as the intersection of two planes, and any normal to either plane must be normal to a vector parallel to their line of intersection. Thus a vector parallel to the line may be obtained as the “cross product” of normals to the two different planes, provided that vector is not 0 — which happens when and only when the two planes are parallel and, therefore, do not determine a line of intersection.

One finds:

(3, −4, 2) × (9, −5, −3) = (22, 27, 21) .

Hence, the required plane has an equation of the form

22x + 27y + 21z = constant ,

and the constant is determined by the fact that the given point must satisfy the equation:

22 · − 1 + 27 · 3 + 21 · 2 = −22 + 81 + 42 = 101 .

Hence, the required plane has the equation:

22x + 27y + 21z = 101 .

8. Find the centroid of the solid cone

x^{2} + y^{2} ≤ z^{2} , 0 ≤ z ≤ 2 .

Response: The lateral boundary of the solid cone, given by x^{2} + y^{2} = z^{2}, is the surface obtained by rotating the line z = y in the plane x = 0 about the z-axis. Thus, for z ≥ 0 the section of the solid cone by the plane normal to the z-axis through the point (0, 0, z) is the disk of radius z with center (0, 0, z). The radius of this (inverted) cone’s base, as well as its altitude, is 2, and, therefore its volume is

\frac{1}{3}πa^{2}h = \frac{8π}{3} .

By symmetry its centroid lies on the z-axis, and the only issue is what is its z-coordinate:

z̄ = \frac{1}{volume} ∫ ∫ ∫ z dV

= \frac{3}{8π} ∫_{0}^{2} zdz ∫ ∫_{x^{2}+y^{2}≤z^{2}} dxdy

= \frac{3}{8π} ∫_{0}^{2} z · (Area of disk of radius z) dz

= \frac{3}{8π} ∫_{0}^{2} πz^{3}dz

= \frac{3}{8} · 4

= \frac{3}{2}

Hence, the centroid of the cone is the point (0, 0, 3π/2).

9. Find the surface integral of the vector field F over the sphere, oriented by its outer normal, of radius 3 with center at the origin when F is given by

F (x, y, z) = (x^{3}, y^{3}, z^{3}) .

Response: The vector field is well-behaved everywhere, particularly in the ball surrounded by the given sphere. The divergence theorem may be applied.

(divF)(x, y, z) = (\frac{∂}{∂x}, \frac{∂}{∂y} , \frac{∂}{∂z}) · F = \frac{∂}{∂x}(x^{3}) + \frac{∂}{∂y} (y^{3}) + \frac{∂}{∂z} (z^{3}) = 3(x^{2} + y^{2} + z^{2}) .

Therefore,

∫ ∫_{S} F · Ndσ = ∫ ∫ ∫_{B} divF dV .

In polar coordinates, observing that dV = ρ^{2} sin ϕ dρdϕdθ and that divF = 3ρ^{2}, the triple integral with radius a = 3 becomes:

3 ∫_{0}^{2π} dθ ∫_{0}^{π} sin ϕdϕ ∫_{0}^{3} ρ^{4} dρ ,

which evaluated becomes

\frac{12π}{5} a^{5} = \frac{2916π}{5} .

10. For a conservative vector field in the plane what can be said about its integral over the circle, traversed counterclockwise, of radius 2 with center at the point (3, 4)? Explain your answer.

Response: Because conservative in this context means that the vector field is the gradient of a scalar, one may use the formula

∫_{C} ∇f = f(B) − f(A)

where A and B are, respectively, the initial and final points of an oriented curve C. A trip around a circle, no matter where it begins, will end where it begins. Thus, A = B, and the integral of a vector field that is conservative in the plane around any circle must be zero.