Math 113 Calculus III Final Exam Practice Problems Spring 2003

Solution:

To find the equation of the plane, we can first find two vectors in the plane:

v = (0, 1, 0) - (1, 0, 0) = (-1, 1, 0)

v = (0, 0, 1) - (1, 0, 0) = (-1, 0, 1)

The normal vector to the plane is the cross product of these two vectors:

n = v v = (11 - 00, 0(-1) - (-1)1, (-1)0 - 1(-1)) = (1, 1, 1)

Using the point-normal form of the plane equation with point (1, 0, 0):

1(x - 1) + 1(y - 0) + 1(z - 0) = 0

x + y + z = 1

Solution:

First Partial Derivatives:

f/x = 3xy + ycos(xy) + e^(x+y)

f/y = 2xy + xcos(xy) + e^(x+y)

Second Partial Derivatives:

f/x = 6xy - ysin(xy) + e^(x+y)

f/y = 2x - xsin(xy) + e^(x+y)

f/xy = 6xy + cos(xy) - xysin(xy) + e^(x+y)

f/yx = 6xy + cos(xy) - xysin(xy) + e^(x+y)

Solution:

First, compute the gradient of f:

f = (f/x, f/y, f/z) = (yz, xz, 2xyz)

Evaluate the gradient at the point (1, 1, 2):

f(1, 1, 2) = (12, 12, 2112) = (4, 4, 4)

Normalize the direction vector v:

|v| = (2 + 1 + (-2)) = 9 = 3

u = v/|v| = (2/3)i + (1/3)j - (2/3)k

The directional derivative is:

D_u f = f(1, 1, 2) u = 4(2/3) + 4(1/3) + 4(-2/3) = 8/3 + 4/3 - 8/3 = 4/3

Solution:

First, find the intersection points of y = x and y = 2 - x:

x = 2 - x 2x = 2 x = 1 x = 1

The region R can be described as:

R = {(x, y) : -1 x 1, x y 2 - x}

Setting up the double integral:

_{-1}^{1} _{x}^{2-x} (x + y) dy dx

First, integrate with respect to y:

_{-1}^{1} [xy + y/2]_{y=x}^{y=2-x} dx

= _{-1}^{1} [x(2-x) + (2-x)/2 - x(x) - (x)/2] dx

= _{-1}^{1} [2x - x + (4 - 4x + x)/2 - x - x/2] dx

= _{-1}^{1} [2x - 2x + 2 - 2x] dx

= [x - x/2 + 2x - 2x/3]_{-1}^{1}

= (1 - 1/2 + 2 - 2/3) - (1 - 1/2 - 2 + 2/3)

= (1 - 1/2 + 2 - 2/3 - 1 + 1/2 + 2 - 2/3)

= 4 - 4/3 = 8/3

Solution:

The region E can be described as:

E = {(x, y, z) : 0 x 1, 0 y 1 - x, 0 z 1 - x - y}

Setting up the triple integral:

_{0}^{1} _{0}^{1-x} _{0}^{1-x-y} z dz dy dx

First, integrate with respect to z:

_{0}^{1} _{0}^{1-x} [z/2]_{z=0}^{z=1-x-y} dy dx

= _{0}^{1} _{0}^{1-x} (1-x-y)/2 dy dx

Next, integrate with respect to y:

= (1/2) _{0}^{1} _{0}^{1-x} (1 - 2x - 2y + x + 2xy + y) dy dx

= (1/2) _{0}^{1} [(1 - 2x + x)y + xy + y/3]_{y=0}^{y=1-x} dx

= (1/2) _{0}^{1} [(1 - 2x + x)(1 - x) + x(1 - x) + (1 - x)/3] dx

Simplifying and integrating with respect to x:

= (1/2) _{0}^{1} (1 - 3x + 3x - x) dx

= (1/2) [x - 3x/2 + x - x/4]_{0}^{1}

= (1/2) (1 - 3/2 + 1 - 1/4)

= (1/2) (3/4) = 3/8

Solution:

Parameterize the curve: x = t, y = t, 0 t 1

Compute the derivatives: dx/dt = 2t, dy/dt = 3t

Substitute into the line integral:

_{0}^{1} [2(t)(t)(2t) + 3(t)(t)(3t)] dt

= _{0}^{1} [2tt(2t) + 3tt(3t)] dt

= _{0}^{1} [4t + 9t] dt

= _{0}^{1} 13t dt

= [13t/13]_{0}^{1} = 1

Solution:

Parameterize the surface using polar coordinates: x = r cos, y = r sin, z = r

The surface can be described as: 0 r 2, 0 2

Compute the surface element:

dS = |r_r r_| dr d

where r(r,) = (r cos, r sin, r)

Compute the partial derivatives:

r_r = (cos, sin, 2r)

r_ = (-r sin, r cos, 0)

Calculate the cross product:

r_r r_ = (-2r cos, -2r sin, r)

Find its magnitude:

|r_r r_| = (4r cos + 4r sin + r) = (4r + r) = r(4r + 1)

Set up the surface integral:

_{0}^{2} _{0}^{2} r r(4r + 1) dr d

= _{0}^{2} d _{0}^{2} r(4r + 1) dr

= 2 _{0}^{2} r(4r + 1) dr

Using substitution u = 4r + 1, du = 8r dr, r = (u-1)/4, r dr = r r dr = (u-1)/4 du/8 = (u-1)du/32:

= 2 _{1}^{17} (u-1)u du/32

= (/16) _{1}^{17} (u^(3/2) - u^(1/2)) du

= (/16) [2/5 u^(5/2) - 2/3 u^(3/2)]_{1}^{17}

= (/16) [(2/517^(5/2) - 2/317^(3/2)) - (2/5 - 2/3)]

= (/16) [2/517^(5/2) - 2/317^(3/2) + 4/15]

= /4017^(5/2) - /2417^(3/2) + /60

= /120 (317^(5/2) - 517^(3/2) + 2)

Solution:

Green's Theorem states:

_C P dx + Q dy = _D (Q/x - P/y) dA

Compute the partial derivatives:

Q/x = /x (10x + cos(y)) = 10

P/y = /y (3y + e^x) = 3

Set up the double integral over the rectangular region D: 0 x 3, 0 y 2:

_D (10 - 3) dA = _{0}^{3} _{0}^{2} 7 dy dx

= 7 3 2 = 42

Solution:

Stokes' Theorem states:

_S ( F) dS = _S F dr

The boundary S is the curve where the paraboloid intersects the xy-plane (z = 0), so:

0 = 1 - x - y x + y = 1

This is a circle of radius 1 in the xy-plane, which can be parameterized as:

x = cos(t), y = sin(t), z = 0, 0 t 2

Evaluating F on this curve (where z = 0):

F(cos(t), sin(t), 0) = (0, 0, cos(t)sin(t))

Compute dr:

dr = (-sin(t), cos(t), 0) dt

Calculate the line integral:

_S F dr = _{0}^{2} (0, 0, cos(t)sin(t)) (-sin(t), cos(t), 0) dt

= _{0}^{2} 0 dt = 0

Therefore, _S ( F) dS = 0

Solution:

The Divergence Theorem states:

_S F dS = _E F dV

Compute the divergence of F:

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

Set up the triple integral over the sphere E: x + y + z 4

Using spherical coordinates:

0 2, 0 2, 0

_E F dV = 3 _0^{2} _0^ _0^2 () sin() d d d

= 3 _0^{2} d _0^ sin() d _0^2 d

= 3 2 2 32/5 = 384/5