Calculus III Practice Final Exam Solutions

Solution:

Dot product:
a b = (2)(-1) + (-1)(4) + (3)(-2) = -2 - 4 - 6 = -12

Cross product:
a b = |i j k|
|2 -1 3|
|-1 4 -2|
= i[(-1)(-2) - (3)(4)] - j[(2)(-2) - (3)(-1)] + k[(2)(4) - (-1)(-1)]
= i[2 - 12] - j[-4 + 3] + k[8 - 1]
= -10i + 1j + 7k = -10, 1, 7

Angle between vectors:
cos() = (a b) / (|a| |b|)
|a| = (2 + (-1) + 3) = (4 + 1 + 9) = 14
|b| = ((-1) + 4 + (-2)) = (1 + 16 + 4) = 21
cos() = -12 / (14 21) = -12/294 -0.70
= arccos(-0.70) 134.4

Solution:
The given plane has normal vector n = 5, -2, 3. Since our desired plane is parallel, it will have the same normal vector.

Using the point-normal form of a plane: a(x - x) + b(y - y) + c(z - z) = 0

Substituting our values:
5(x - 2) - 2(y - 1) + 3(z + 1) = 0
5x - 10 - 2y + 2 + 3z + 3 = 0
5x - 2y + 3z - 5 = 0

Therefore, the equation of the plane is 5x - 2y + 3z = 5.

Solution:

First-order partial derivatives:
f = /x (xy + e^(xy)) = 2xy + ye^(xy)
f_y = /y (xy + e^(xy)) = x + xe^(xy)

Second-order partial derivatives:
f = /x f(x,y) = /x (2xy + ye^(xy)) = 2y + ye^(xy)
f_yy = /y f(x,y) = /y (x + xe^(xy)) = xe^(xy)
fy = /yx f(x,y) = /y (2xy + ye^(xy)) = 2x + e^(xy) + xye^(xy)
f_y = /xy f(x,y) = /x (x + xe^(xy)) = 2x + e^(xy) + xye^(xy)

Note that fy = f_y as expected from Clairaut's Theorem.

Solution:
First, find the intersection points of the two curves:
x = 2x => x - 2x = 0 => x(x - 2) = 0
This gives x = 0 and x = 2.

For 0 x 2, we have x 2x.
Therefore, the region R is: {(x,y) | 0 x 2, x y 2x}

_R (x + y) dA = _{x}^{2x} (x + y) dy dx

First, integrate with respect to y:
_{x}^{2x} (x + y) dy = [xy + y]_{x}^{2x}
= x(2x) + (2x) - x(x) - (x)
= 2x + 2x - x - x
= 4x - x - x

Now, integrate with respect to x:
(4x - x - x) dx = [4/3x - x - 1/10x]
= (4/3)(8) - (16) - 1/10(32) - 0
= 32/3 - 4 - 32/10
= 32/3 - 4 - 16/5
= (160 - 60 - 48)/15
= 52/15

Solution:
In cylindrical coordinates:
x = r cos(), y = r sin(), z = z
x + y = r
dV = r dz dr d

The cylinder x + y = 4 becomes r = 4, so r = 2.
The bounds are:
0 r 2
0 2
0 z x + 2 = r cos() + 2

_E z dV = ^{2} ^{r cos()+2} z r dz dr d

First, integrate with respect to z:
^{r cos()+2} z r dz = r [z]^{r cos()+2}
= r (r cos() + 2)
= r(r cos() + 4r cos() + 4)
= r cos() + 2r cos() + 2r

Now, integrate with respect to r:
(r cos() + 2r cos() + 2r) dr
= [1/8r cos() + 2/3r cos() + r]
= 1/8(16) cos() + 2/3(8) cos() + 4
= 2 cos() + 16/3 cos() + 4

Finally, integrate with respect to :
^{2} (2 cos() + 16/3 cos() + 4) d
= ^{2} (2( + cos(2)) + 16/3 cos() + 4) d
= ^{2} (1 + cos(2) + 16/3 cos() + 4) d
= ^{2} (5 + cos(2) + 16/3 cos()) d
= [5 + sin(2) + 16/3 sin()]^{2}
= 5(2) + sin(4) + 16/3 sin(2) - 0
= 10

Solution:
Parameterize C using x = 2 cos(), y = 2 sin() for -/2 /2.

Calculate ds:
x' = -2 sin(), y' = 2 cos()
ds = [(x') + (y')] d = [4 sin() + 4 cos()] d = [4(sin() + cos())] d = 2 d

Substitute into the integral:
_C xy^4 ds = _{-/2}^{/2} (2 cos())(2 sin())^4 2 d
= _{-/2}^{/2} 2 cos() 16 sin() 2 d
= 64 _{-/2}^{/2} cos() sin() d

Let u = sin(), du = cos() d
When = -/2, u = -1
When = /2, u = 1

= 64 _{-1}^1 u^4 du
= 64 [1/5 u^5]_{-1}^1
= 64/5 (1 - (-1))
= 128/5

Solution:
Write the surface as z = f(x,y) = 2 + 3x + 4y

For a surface given by z = f(x,y):
dS = [f_x + f_y + 1] dA

Calculate partial derivatives:
f_x = 3, f_y = 4
dS = [3 + 4 + 1] dA = [9 + 16 + 1] dA = 26 dA

Set up the integral:
_S x dS = _D x 26 dA
where D is the rectangle [0,1] [0,2]
= 26 x dy dx
= 26 x [y] dx
= 26 2x dx
= 226 [1/3 x]
= 226 1/3
= 226/3

Solution:
Green's Theorem states: _C P dx + Q dy = _D (Q/x - P/y) dA

Here, P = y + 2x and Q = 3y - 4x
Q/x = -8x and P/y = 2y

_C (y + 2x) dx + (3y - 4x) dy = _D (-8x - 2y) dA

The region D is enclosed by y = x and y = 4.
Finding the intersection points: x = 4, so x = -2 and x = 2.

Setting up the integral: _{-2}^2 _{x^2}^4 (-8x - 2y) dy dx

First, integrate with respect to y:
_{x}^4 (-8x - 2y) dy = [-8xy - y]_{x}^4
= -8x(4) - 4 - (-8x(x) - (x))
= -32x - 16 + 8x + x
= x + 8x - 32x - 16

Now, integrate with respect to x:
_{-2}^2 (x + 8x - 32x - 16) dx
= [1/5 x + 2x - 16x - 16x]_{-2}^2
= (1/5(32) + 2(16) - 16(4) - 16(2)) - (1/5(-32) + 2(16) - 16(4) + 16(2))
= (32/5 + 32 - 64 - 32) - (-32/5 + 32 - 64 + 32)
= (32/5 - 64) - (-32/5)
= 32/5 - 64 + 32/5
= 64/5 - 64
= -256/5

Solution:
Stokes' Theorem states: _S curl F dS = _C F dr

By Stokes' Theorem, the flux of curl F through S equals the line integral of F around the boundary curve C.
The boundary curve C is the intersection of the sphere x + y + z = 4 and the cylinder x + y = 1 above the xy-plane.

Find the intersection:
At the intersection, x + y = 1, so from the sphere equation:
1 + z = 4, meaning z = 3, z = 3 (since we want the part above the xy-plane)

So C is the circle x + y = 1 at height z = 3.
Parameterize C as: r(t) = cos(t), sin(t), 3 for 0 t 2

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

Evaluate F at C:
F(r(t)) = -sin(t), cos(t), (3) = -sin(t), cos(t), 3

Calculate F dr:
F dr = -sin(t), cos(t), 3 -sin(t), cos(t), 0 dt
= sin(t) + cos(t) dt
= 1 dt

Therefore:
_C F dr = _0^{2} 1 dt = [t]_0^{2} = 2

By Stokes' Theorem, _S curl F dS = 2.

Solution:
The Divergence Theorem states: _S F dS = _E div F dV

Calculate the divergence:
div F = /x (xy) + /y (yz) + /z (zx)
= y + z + x

Apply the Divergence Theorem:
_S F dS = _E (x + y + z) dV
where E = [0,1] [0,2] [0,3]

Set up the triple integral:
(x + y + z) dz dy dx

First, integrate with respect to z:
(x + y + z) dz = [xz + yz + z]
= 3x + 3y + 9

Next, integrate with respect to y:
(3x + 3y + 9) dy = [3xy + y + 9y]
= 6x + 8 + 18
= 6x + 26

Finally, integrate with respect to x:
(6x + 26) dx = [2x + 26x]
= 2 + 26
= 28

Therefore, _S F dS = 28.