Mathematics 241 – Multivariable Calculus Midterm 3 – Solutions November 30

I.

A) (15) Let F be the vector field

F(x, y, z) = (xz − y^{2}, xy + z, 2x + sin(y)).

Compute the divergence div(F) and the curl curl(F).

Solution: Using the ∇ operator, the divergence is

div(F) = ∇ · F = \frac{∂}{∂x}(xz − y^{2}) + \frac{∂}{∂y} (xy + z) + \frac{∂}{∂z} (2x + sin(y)) = z + x.

The curl is

curl(F) = ∇ × F = (cos(y) − 1, x − 2, 3y).

B) (5) Show that if F = (F_{1}, F_{2}, F_{3}) is any vector field on a domain in R^{3} whose component functions have continuous second-order partial derivatives, then div(curl(F)) = 0.

Solution:

div(curl(F)) = div (\frac{∂F_{3}}{∂y} − \frac{∂F_{2}}{∂z} , − (\frac{∂F_{3}}{∂x} − \frac{∂F_{1}}{∂z} ) , \frac{∂F_{2}}{∂x} − \frac{∂F_{1}}{∂y} )
= \frac{∂}{∂x} (\frac{∂F_{3}}{∂y} − \frac{∂F_{2}}{∂z} ) + \frac{∂}{∂y} (− (\frac{∂F_{3}}{∂x} − \frac{∂F_{1}}{∂z} )) + \frac{∂}{∂z} (\frac{∂F_{2}}{∂x} − \frac{∂F_{1}}{∂y} )
= \frac{∂^{2}F_{3}}{∂y∂x} − \frac{∂^{2}F_{2}}{∂z∂x} − \frac{∂^{2}F_{3}}{∂x∂y} + \frac{∂^{2}F_{1}}{∂z∂y} + \frac{∂^{2}F_{2}}{∂x∂z} − \frac{∂^{2}F_{1}}{∂y∂z}
= 0

by the equality of the mixed partial derivatives.

II. All parts of this problem refer to the function

f(x, y) = x^{3} + y^{2} − 6xy + 6x + 3y.

A) (10) Determine all critical points of f.

Solution: The critical points are the solutions of

0 = f_{x} = 3x^{2} − 6y + 6
0 = f_{y} = 2y − 6x + 3

Solving for y in the second equation, y = 3x−3/2. Substituting into the first equation, 0 = 3x^{2} − 18x + 15 = 3(x − 1)(x − 5) and x = 1 or x = 5. The critical points are

(1, 3/2), (5, 27/2).

B) (15) Use the Hessian Criterion (second derivative test) to classify each of the critical points as a local maximum, a local minimum, a saddle point, or a degenerate critical point.

Solution: The second derivatives of f are

f_{xx} = 6x, f_{xy} = −6, f_{yy} = 2.

• At (1, 3/2), the Hessian matrix is

D^{2}(f)(1, 3/2) = (\begin{matrix}6 & −6 \\ −6 & 2 \end{matrix})

Since det(D^{2}(f)(1, 3/2)) = 12 − 36 = −24 < 0, this is a saddle point.

• At (5, 27/2), the Hessian matrix is

D^{2}(f)(5, 27/2) = (\begin{matrix}30 & −6 \\ −6 & 2 \end{matrix})

Since det(D^{2}(f)(5, 7/2)) = 60 − 36 = 24 > 0 and 30 > 0, this is a local minimum.

III. (20) A mountain has the shape of the graph z = f(x, y) = 4 − x^{2} − \frac{y^{2}}{4} , where x, y are east-west and north-south map coordinates respectively, and z is the altitude (in thousands of feet). If you hike along a trail on the mountain described by the circle g(x, y) = x^{2}+y^{2} = 1 on the map, what are the highest and lowest altitudes that you reach?

Solution: We want to maximize and minimize f(x, y) on the constraint curve g(x, y) = 1, so this is a job for Lagrange multipliers:

f_{x} = −2x = 2λx = λg_{x}
f_{y} = −y/2 = 2λy = λg_{y}
x^{2} + y^{2} = 1

From the x equation, we see that either x = 0 or λ = −1. If x = 0, then y = ±1 from the constraint equation. If λ = −1, then the y equation gives y = 0, and x = ±1 from the constraint. Hence there are four points that solve the LM equations: (1, 0), (−1, 0), (0, 1), (0, −1). The altitudes at these points are f(1, 0) = f(−1, 0) = 3 (thousand feet) and f(0, 1) = f(0, −1) = 3.75 (thousand feet). The maximum altitude is 3750 and the minimum altitude is 3000 feet at the points on the trail.

IV. All parts of this problem refer to the region D in the right half-plane (x ≥ 0), bounded by y = x, y = −x and x^{2} + y^{2} = 9.

A) (5) Sketch the region D.

Solution: The region is a quarter circle, with the straight edges along the two given lines and curved edge along the circle of radius 3, center (0, 0).

B) (10) What would the limits of integration be to compute ∫ ∫_{D} f(x, y) dx dy integrating with respect to y first, then x? Solution: This way, we must split the region along x = 3/\sqrt{2}.

∫_{0}^{3/\sqrt{2}} ∫_{−x}^{x} f(x, y) dydx + ∫_{3/\sqrt{2}}^{3} ∫_{−\sqrt{9−x^{2}}}^{√\overline{9−x^{2}}} f(x, y) dydx.

C) (10) What would the limits of integration be to compute ∫ ∫_{D} f(x, y) dx dy integrating with respect to x first, then y? Solution: We must split the region along the line y = 0

∫_{−3/\sqrt{2}}^{0} ∫_{−y}^{√\overline{9−y^{2}}} f(x, y) dxdy + ∫_{0}^{3/\sqrt{2}} ∫_{y}^{√\overline{9−y^{2}}} f(x, y) dxdy.

D) (10) Using any applicable method, compute ∫ ∫_{D} x dx dy. Solution 1: Changing to polar coordinates this becomes

∫_{−π/4}^{π/4} ∫_{0}^{3}(r cos θ)r drdθ = ∫_{0}^{3} r^{2} dr · ∫_{−π/4}^{π/4} cos θ dθ
= \frac{r^{3}}{3} |_{0}^{3} · sin θ|_{−π/4}^{π/4}
= 9\sqrt{2}.

Solution 2: If we use the form from B above, the first integral is

∫_{0}^{3/\sqrt{2}} ∫_{−x}^{x} x dydx = ∫_{0}^{3/\sqrt{2}} xy|_{y=−x}^{y=x} dx
= ∫_{0}^{3/\sqrt{2}} 2x^{2} dx
= \frac{2x^{3}}{3} |_{0}^{3/\sqrt{2}}
= \frac{9\sqrt{2}}{2}.

The second integral is evaluated using a u-subsitution after the first integration (the integral from the table is not needed, or even applicable):

∫_{3/\sqrt{2}}^{3} ∫_{−\sqrt{9−x^{2}}}^{√\overline{9−x^{2}}} x dydx = ∫_{3/\sqrt{2}}^{3} 2x\sqrt{9 − x^{2}} dx
= −\frac{2}{3}(9 − x^{2})^{3/2}|_{3/sqrt2}^{3}
= \frac{2}{3}(9/2)^{3/2}
= \frac{2}{3} \frac{3^{3}}{2\sqrt{2}} = \frac{9\sqrt{2}}{2}.

Then the integral is the sum

\frac{9\sqrt{2}}{2} + \frac{9\sqrt{2}}{2} = 9\sqrt{2}.

Extra Credit (10) Compute the triple integral of the function f(x, y, z) = e^{(x^{2}+y^{2}+z^{2})^{3/2}} over the unit ball x^{2} + y^{2} + z^{2} ≤ 1 in R^{3}. (A good choice of coordinates is essential here!)

Solution: Spherical coordinates are the best choice. The integral to be computed is

∫_{0}^{2π} ∫_{0}^{π} ∫_{0}^{1} e^{ρ^{3}}ρ^{2} sin ϕ dρ dϕ dθ.

Since the function is a product of functions of ρ, ϕ, and θ we can use the “product trick”:

= 2π ∫_{0}^{π} sin ϕ dϕ ∫_{0}^{1} ρ^{2}e^{ρ^{3}} dρ = 4π ∫_{0}^{1} ρ^{2}e^{ρ^{3}} dρ.

For the last integral here, let u = ρ^{3}, then du = 3ρ^{2}dρ and the integral is just ∫ e^{u}du up to a constant. The final answer is

\frac{4π}{3} (e − 1).