Problem 1: Partial Derivatives and Gradient

Let f(x,y,z) = xy + yz - xyz. Find:

  1. The partial derivatives fx, fy, and fz
  2. The gradient vector f at point (1, -1, 2)
  3. The directional derivative of f at point (1, -1, 2) in the direction of vector v = 1, 2, -2

Solution:

Part 1: Finding Partial Derivatives

To find the partial derivative of f with respect to x, we treat y and z as constants:

fx = /x(xy + yz - xyz) = 2xy - yz

For the partial derivative with respect to y, treating x and z as constants:

fy = /y(xy + yz - xyz) = x + z - xz

For the partial derivative with respect to z, treating x and y as constants:

fz = /z(xy + yz - xyz) = 2yz - xy

Part 2: Finding the Gradient

The gradient f = fx, fy, fz. Evaluating at point (1, -1, 2):

fx(1, -1, 2) = 2(1)(-1) - (-1)(2) = 0
fy(1, -1, 2) = (1) + (2) - (1)(2) = 3
fz(1, -1, 2) = 2(-1)(2) - (1)(-1) = -3

Therefore, f(1, -1, 2) = 0, 3, -3

Part 3: Finding the Directional Derivative

First, we find the unit vector in the direction of v = 1, 2, -2:

|v| = (1 + 2 + (-2)) = (1 + 4 + 4) = 9 = 3
u = v/|v| = 1/3, 2/3, -2/3

The directional derivative is:

Duf(1, -1, 2) = f(1, -1, 2) u = 0, 3, -3 1/3, 2/3, -2/3
= 0(1/3) + 3(2/3) + (-3)(-2/3) = 2 + 2 = 4

Problem 2: Double Integration

Evaluate the double integral (x + y) dA over the region D bounded by y = x, y = 2x, and x = 1.

Solution:

First, let's sketch the region D. It's bounded by two rays y = x and y = 2x, and the vertical line x = 1. This region is best integrated dy dx:

(x + y) dA = (x + y) dy dx

Solving the inner integral:

(x + y) dy = [xy + y/3]
= x(2x) + (2x)/3 - x(x) - x/3
= 2x + 8x/3 - x - x/3 = x + 7x/3 = 10x/3

Now solving the outer integral:

(10x/3) dx = [10x/12] = 10/12 = 5/6

Therefore, (x + y) dA = 5/6 over the given region.

Problem 3: Triple Integrals in Cylindrical Coordinates

Evaluate the triple integral E z dV where E is the solid bounded by the paraboloid z = x + y and the plane z = 9.

Solution:

This problem is best solved using cylindrical coordinates. In cylindrical coordinates:

x = r cos(), y = r sin(), z = z
dV = r dz dr d

The paraboloid becomes z = r in cylindrical coordinates, and our region E is between z = r and z = 9.

To find the limits of r, we need to know where the paraboloid intersects the plane:

r = 9 r = 3

Therefore, r ranges from 0 to 3, and ranges from 0 to 2.

Setting up the triple integral:

E z dV = z r dz dr d

Solving the inner integral:

z r dz = r[z/2] = r(81/2 - r/2)

Solving the middle integral:

r(81/2 - r/2) dr = (81r/2 - r/2) dr
= [81r/4 - r/12] = 729/6 = 121.5

Finally, solving the outer integral:

121.5 d = 121.5[] = 121.5(2) = 243

Therefore, E z dV = 243.

Problem 4: Line Integrals

Compute the line integral C xy dx + xy dy where C is the positively oriented triangle with vertices (0,0), (2,0), and (2,1).

Solution:

The triangle C consists of three line segments:

  • C: From (0,0) to (2,0)
  • C: From (2,0) to (2,1)
  • C: From (2,1) back to (0,0)

Over C (y = 0, dy = 0, x from 0 to 2):

C xy dx + xy dy = x0 dx + x00 = 0 dx = 0

Over C (x = 2, dx = 0, y from 0 to 1):

C xy dx + xy dy = 2y0 + 2y dy = 2y dy = [2y/3] = 2/3

Over C (line from (2,1) to (0,0)):

The equation of the line is y = x/2, so dy = dx/2. Since we're going from (2,1) to (0,0), x goes from 2 to 0.

C xy dx + xy dy = x(x/2) dx + x(x/2)(dx/2)
= x/2 dx + x/8 dx
= (5x/8) dx = [5x/32]
= 0 - 5(2)/32 = -80/32 = -5/2

Total line integral:

C = C + C + C = 0 + 2/3 - 5/2 = 4/6 - 15/6 = -11/6

Therefore, C xy dx + xy dy = -11/6.

Problem 5: Surface Integrals

Compute the surface integral S yz dS where S is the part of the plane x + y + z = 1 that lies in the first octant.

Solution:

For a surface given by z = g(x,y), the surface integral can be expressed as:

S f(x,y,z) dS = D f(x,y,g(x,y))(1 + (g/x) + (g/y)) dA

For our surface, we have z = g(x,y) = 1 - x - y, so:

g/x = -1 and g/y = -1

This gives:

(1 + (g/x) + (g/y)) = (1 + 1 + 1) = 3

So our surface integral becomes:

S yz dS = D y(1 - x - y)3 dA

Now we need to determine the region D in the xy-plane. Since S lies in the first octant and is bounded by x + y + z = 1, we have x 0, y 0, and z = 1 - x - y 0, which means x + y 1.

Setting up the double integral:

D y(1 - x - y)3 dA = 3 y(1 - x - y) dy dx

Solving the inner integral:

y(1 - x - y) dy = (y - xy - y) dy
= [y/2 - xy/2 - y/3] = (1-x)/6

Solving the outer integral:

3 (1-x)/6 dx = 3/6 u(-du) [using substitution u = 1-x]
= 3/6 [u/4] = 3/6 1/4 = 3/24

Therefore, S yz dS = 3/24.

Problem 6: Divergence Theorem

Use the Divergence Theorem to calculate S F dS where F = x, y, z and S is the surface of the ball x + y + z = 4.

Solution:

The Divergence Theorem states that S F dS = E div(F) dV, where E is the solid bounded by S.

First, let's compute the divergence of F:

div(F) = /x(x) + /y(y) + /z(z) = 3x + 3y + 3z = 3(x + y + z)

The region E is the ball x + y + z 4, which is a sphere of radius 2. This problem is best solved using spherical coordinates:

x = sin() cos(), y = sin() sin(), z = cos()
dV = sin() d d d
x + y + z =

In spherical coordinates, the divergence becomes div(F) = 3

Setting up the triple integral:

E div(F) dV = 3 sin() d d d = 3 sin() d d d

Solving the inner integral:

3 d = [3/5] = 3(2)/5 = 96/5

Solving the middle integral:

96/5 sin() d = 96/5[-cos()] = 96/52 = 192/5

Solving the outer integral:

192/5 d = 192/5[] = 192/5 2 = 384/5

Therefore, by the Divergence Theorem, S F dS = 384/5.

Problem 7: Stokes' Theorem

Use Stokes' Theorem to evaluate S curl(F) dS where F = y, -x, z and S is the part of the sphere x + y + z = 4 that lies above the plane z = 1, oriented upward.

Solution:

Stokes' Theorem relates the surface integral of the curl of a vector field to a line integral around the boundary curve C:

S curl(F) dS = C F dr

First, let's find the boundary curve C of our surface. The surface is the part of the sphere x + y + z = 4 that lies above the plane z = 1. The boundary curve is the intersection of the sphere and the plane:

x + y + 1 = 4 x + y = 3

This is a circle of radius 3 in the plane z = 1, centered on the z-axis.

We can parameterize this curve C as:

r(t) = 3 cos(t), 3 sin(t), 1 where 0 t 2

The derivative of this parameterization is:

r'(t) = -3 sin(t), 3 cos(t), 0

Now, let's find F(r(t)):

F(r(t)) = 3 sin(t), -3 cos(t), 1

The line integral becomes:

C F dr = F(r(t)) r'(t) dt
= (3 sin(t))(-3 sin(t)) + (-3 cos(t))(3 cos(t)) + 1(0) dt
= -3 sin(t) - 3 cos(t) dt
= -3(sin(t) + cos(t)) dt
= -3 dt = -6

Therefore, by Stokes' Theorem, S curl(F) dS = -6.

Problem 8: Lagrange Multipliers

Find the maximum and minimum values of f(x,y) = x + y subject to the constraint g(x,y) = x + 2y = 3 using the method of Lagrange multipliers.

Solution:

To find the extreme values of f subject to the constraint g = 3 using Lagrange multipliers, we solve the system of equations:

f = g and g(x,y) = 3

First, let's compute the gradients:

f = 2x, 2y and g = 2x, 4y

Setting up the Lagrange equations:

2x = (2x), 2y = (4y), x + 2y = 3

From the first equation, either x = 0 or = 1.

Case 1: x = 0

If x = 0, then from our constraint: y = 3/2 y = (3/2)

From the second Lagrange equation with x = 0: 2y = (4y), which gives = 1/2 when y 0.

So we have points (0, (3/2)) and (0, -(3/2)).

Evaluating f at these points: f(0, (3/2)) = 0 + (3/2) = 3/2

Case 2: = 1

If = 1, then from the second Lagrange equation: 2y = 4y y = 0

From our constraint with y = 0: x = 3 x = 3

So we have points (3, 0) and (-3, 0).

Evaluating f at these points: f(3, 0) = 3 + 0 = 3

Now, let's determine which are maxima and which are minima:

  • f(0, (3/2)) = 3/2 (minimum)
  • f(3, 0) = 3 (maximum)

Therefore, the function f(x,y) = x + y has a minimum value of 3/2 at points (0, (3/2)) and (0, -(3/2)), and a maximum value of 3 at points (3, 0) and (-3, 0) under the constraint x + 2y = 3.