Mathematics 241 – Multivariable Calculus Solutions for Exam 3 Review Problems November 28, 2007

I. Let F(x, y, z) = (x + yz, y + xz, z + xy) be a vector field on R^{3}.

A) Compute the divergence of F.

Solution: The divergence of a vector field F is the scalar-valued function div(F) = ∇·F. Here:

(\frac{∂}{∂x}, \frac{∂}{∂y} , \frac{∂}{∂z} ) · (x + yz, y + xz, z + xy) = 3.

B) Compute the curl of F.

Solution: The curl of a vector field F is the vector field curl(F) = ∇ × F (cross product). Here:

(\frac{∂}{∂x}, \frac{∂}{∂y} , \frac{∂}{∂z} ) × (x + yz, y + xz, z + xy) = (0, 0, 0).

C) Is there a scalar-valued function f(x, y, z) such that F = ∇f? Why or why not?

Solution: For a scalar-valued function, ∇f is the gradient vector field. There are such functions:

f = \frac{1}{2}(x^{2} + y^{2} + z^{2}) + xyz + C

where C is an arbitrary constant.

II. All parts of this question refer to the function

f(x, y) = xe^{−x^{2}} cos(y).

A) Compute the second degree Taylor polynomial of f at (a, b) = (0, 0). Give both the form with the derivative and Hessian matrices and the expanded (“messy, practical”) form.

Solution: The form with the derivative and Hessian matrices is

p_{2}(x, y) = f(0, 0) + D(f)(0, 0) ( \begin{matrix}x − 0 \\ y − 0 \end{matrix} ) + \frac{1}{2}(x − 0 y − 0)D^{2}(f) ( \begin{matrix}x − 0 \\ y − 0 \end{matrix} ) .

Expanded out this becomes

\begin{matrix}f(0, 0)+ & \frac{∂f}{∂x} (0, 0)(x − 0) + \frac{∂f}{∂x} (0, 0)(x − 0) \\ + & \frac{1}{2} (\frac{∂^{2}f}{∂x^{2}} (0, 0)(x − 0)^{2} + 2\frac{∂^{2}f}{∂x∂y} (0, 0)(x − 0)(y − 0) + \frac{∂^{2}f}{∂y^{2}} (0, 0)(y − 0)^{2}) . \end{matrix}

For our function f(x, y),

\frac{∂f}{∂x} = cos(y)e^{−x^{2}}(1 − 2x^{2}) ⇒ \frac{∂f}{∂x} (0, 0) = 1
\frac{∂f}{∂y} = − sin(y)xe^{−x^{2}} ⇒ \frac{∂f}{∂x} (0, 0) = 0
\frac{∂^{2}f}{∂x^{2}} = cos(y)e^{−x^{2}}(−6x + 4x^{3}) ⇒ \frac{∂^{2}f}{∂x^{2}} (0, 0) = 0
\frac{∂^{2}f}{∂x∂y} = − sin(y)e^{−x^{2}}(1 − 2x^{2}) ⇒ \frac{∂^{2}f}{∂x∂y} (0, 0) = 0
\frac{∂^{2}f}{∂y^{2}} = − cos(y)xe^{−x^{2}} ⇒ \frac{∂f}{∂x} (0, 0) = 0

Hence

p_{2}(x, y) = x.

B) f(x, y) has exactly three critical points with 0 ≤ y ≤ 2π/3. Find them.

Solution: The critical points are the simultaneous solutions of \frac{∂f}{∂x} = 0 and \frac{∂f}{∂y} = 0. From the computations of the partial derivatives in part A, for y in the range given,

\frac{∂f}{∂x} = cos(y)e^{−x^{2}}(1 − 2x^{2}) = 0

implies y = π/2 or x = ±1/\sqrt{2}. Then

\frac{∂f}{∂y} = − sin(y)xe^{−x^{2}} = 0

implies y = 0 or x = 0. Putting these together, we find the three critical points:

(0, π/2), (1/\sqrt{2}, 0), (−1/\sqrt{2}, 0).

C) Classify the critical points from part B as local maxima, local minima, or saddle points using the Second Derivative Test (Hessian Criterion). Solution: Using the partial derivatives computed in part A,

D^{2}(f) = ( \begin{matrix}cos(y)e^{−x^{2}}(−6x + 4x^{3}) & − sin(y)e^{−x^{2}}(1 − 2x^{2}) \\ − sin(y)e^{−x^{2}}(1 − 2x^{2}) & − cos(y)xe^{−x^{2}} \end{matrix} )

• At (0, π/2),

D^{2}(f)(0, π/2) = ( \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} ) .

The determinant is det(D^{2}(f)) = −1, so this is a saddle point.

• At (1/\sqrt{2}, 0),

D^{2}(f)(1/\sqrt{2}, 0) = ( \begin{matrix}−4e^{−1/2}/\sqrt{2} & 0 \\ 0 & −e^{−1/2}/\sqrt{2} \end{matrix} ) .

The upper left entry is negative and the determinant is det(D^{2}(f)) = 2e^{−1} > 0, so this is a local maximum.

• At (−1/\sqrt{2}, 0),

D^{2}(f)(−1/\sqrt{2}, 0) = ( \begin{matrix}4e^{−1/2}/\sqrt{2} & 0 \\ 0 & e^{−1/2}/\sqrt{2} \end{matrix} ) .

The upper left entry is positive and the determinant is det(D^{2}(f)) = 2e^{−1} > 0, so this is a local minimum.

III. Let f(x, y) = (x − 1)^{2} + y^{2} and g(x, y) = x^{2}/4 + y^{2}.

A) Find the largest and smallest values of f on the constraint curve defined by g(x, y) = 1 using the method of Lagrange Multipliers.

Solution: The Lagrange multiplier equations are:

2(x − 1) = λx/2
2y = 2λy
x^{2}/4 + y^{2} − 1 = 0.

From the second equation, y = 0 or λ = 1. In the first case, from the constraint equation, x = ±2. In the second case, from the first equation 2x − 2 = x/2, so x = 4/3 and then y = ±\sqrt{5}/3 from the constraint equation. To decide which points are maxima and which are minima of f on the constraint curve we substitute:

\begin{aligned}f(2, 0) & = 1 \\ f(−2, 0) & = 9 \\ f(4/3, ±\sqrt{5}/3) & = 2/3 \end{aligned}

So the maximum value is 9 and the minimum value is 2/3.

B) Sketch the level curves of f for the values c = 2/3, 1, 9 together with the curve g(x, y) = 1 and explain how your sketch relates to the result of part A.

Solution: The level curves of f for c > 0 are circles with center at (1, 0) and radius \sqrt{c}. The constraint curve g(x, y) = 1 is an ellipse with semimajor axis 2 along the x-axis and semiminor axis 1 along the y-axis. The level curves of f for the values c = 2/3, 1, 9 are tangent to the ellipse at the constrained critical points found in part A.

C) What are the absolute maximum and minimum values of h(x, y) < /EM >= 10 − f(x, y) on the region defined by g(x, y) ≤ 1?

Solution: h(x, y) = 10 − (x − 1)^{2} − y^{2} has a critical point at (x, y) = (1, 0) which is a local and global maximum. Applying Lagrange multipliers to h on the constraint curve gives a system of equations with the same solutions as those from part A. We have h(2, 0) = 9, h(−2, 0) = 1, and h(4/3, ±\sqrt{5}/3) = 28/3. So the maximum of h(x, y) is 10 and the minimum is 1.

IV. All parts of this question refer to the region R in the plane inside the circle x^{2} + y^{2} = 5, below parabola y = x^{2} + 1, and above the x-axis.

A) Set up iterated integral(s) to integrate a general function f(x, y) over R, integrating first with respect to x.

Solution: The parabola crosses the circle at (±1, 2) and extends only down to y = 1. So to integrate first with respect to x (type 2 regions), we actually need to subdivide D into three regions: two with 1 ≤ y ≤ 2 and one with 0 ≤ y ≤ 1:

∫_{1}^{2} ∫_{−\sqrt{5−y^{2}}}^{−\sqrt{y−1}} f(x, y) dx dy + ∫_{1}^{2} ∫\begin{matrix}√ & \overline{y−1} \\ √ & \overline{5−y^{2}} \end{matrix} f(x, y) dx dy + ∫_{0}^{1} ∫_{−\sqrt{5−y^{2}}}^{√\overline{5−y^{2}}} f(x, y) dx dy.

B) Same question as A, but interchanging the order of integration (integrating first with respect to y).

Solution: The setup using type 1 regions is similar since the top edge of the region changes at x = −1 and again at x = 1:

∫_{−\sqrt{5}}^{−1} ∫_{0}^{√\overline{5−x^{2}}} f(x, y) dy dx + ∫_{−1}^{1} ∫_{0}^{x^{2}+1} f(x, y) dy dx + ∫_{1}^{√\overline{5}} ∫_{0}^{√\overline{5−x^{2}}} f(x, y) dy dx.

V.

A) State the change of variables formula for double integrals in general.

Solution: See the text and the class notes. The key points are that if T : R_{u,v}^{2} → R_{x,y}^{2} is a change of coordinate mapping, and T (D∗) = D, then

∫ ∫_{D} f(x, y) dxdy = ∫ ∫_{D∗} f(x(u, v), y(u, v)) |\frac{∂(x, y)}{∂(u, v)}| dudv.

Here

\frac{∂(x, y)}{∂(u, v)} = det ( \begin{matrix}\frac{∂x}{∂u} & \frac{∂x}{∂v} \\ \frac{∂y}{∂u} & \frac{∂y}{∂v} \end{matrix} )

is the Jacobian determinant – the determinant of the derivative matrix of the mapping T.

B) Where does the “extra r” in the polar coordinate integral ∫ ∫ f(r cos θ, r sin θ)r dr dθ come from?

Solution: It comes from the Jacobian determinant of the polar coordinate mapping T (r, θ) = (r cos θ, r sin θ):

\frac{∂(x, y)}{∂(r, θ)} = det ( \begin{matrix}cos θ & −r sin θ \\ sin θ & r cos θ \end{matrix} ) = r(cos^{2} θ + sin^{2} θ) = r.

C) Change variables to polar coordinates to determine the integral of f(x, y) = xy over the quarter circle inside x^{2} + y^{2} = 4 in the second quadrant.

Solution: This is

∫ ∫_{D} xy dxdy = ∫_{π/2}^{π} ∫_{0}^{2}(r cos θ)(r sin θ) rdrdθ

= ∫_{π/2}^{π} ∫_{0}^{2} r^{3} cos θ sin θ drdθ

= \frac{r^{4}}{4} |_{0}^{2} · \frac{1}{2}sin^{2} θ|_{π/2}^{π}

= −2.

D) Change variables to spherical coordinates in R^{3} to evaluate the triple integral of the function x over the portion of the solid ball of radius 1 with center at (0,0,0) in the first octant (i.e. the part of the region inside the sphere of radius 1, with x, y, z all ≥ 0).

Solution: This is

∫_{0}^{π/2} ∫_{0}^{π/2} ∫_{0}^{1} ρ sin ϕ cos θ · ρ^{2} sin ϕ dρ dϕ dθ = \frac{π}{16}.

VI. Find the volume of the solid bounded above by the cone z = (x^{2} + y^{2})^{1/2}, below by z = 0, and by 0 ≤ x^{2} + y^{2} ≤ 4. You may use any convenient coordinate system to carry out the necessary computations.

Solution: Because of the circular symmetry of the region about the z-axis, cylindrical coordinates are indicated for this one. The volume is

V = ∫_{0}^{2π} ∫_{0}^{2} ∫_{0}^{r} 1 rdz dr dθ = \frac{16π}{3} .