Math 211 Multivariable Calculus Final Exam Solutions

To find the equation of a plane passing through three points, we need a normal vector to the plane and a point on the plane. The normal vector can be found by taking the cross product of vectors lying on the plane.

First, find vectors AB and AC:

AB = B - A = (2-1, 0-2, -1-3) = (1, -2, -4)

AC = C - A = (3-1, 1-2, 2-3) = (2, -1, -1)

Now find the cross product n = AB AC:

Computing the determinant:

n = i((-2)(-1) - (-4)(-1)) - j((1)(-1) - (-4)(2)) + k((1)(-1) - (-2)(2))

n = i(2 - 4) - j(-1 + 8) + k(-1 + 4)

n = i(-2) - j(7) + k(3)

n = (-2, -7, 3)

Using point A(1, 2, 3) and the normal vector n, the equation of the plane is:

Rearranging:

-2x + 2 - 7y + 14 + 3z - 9 = 0

-2x - 7y + 3z + 7 = 0

To find the intersection curve, we substitute z = 4 into the equation of the surface:

4 = x + y

In the xy-plane, this is the equation of a circle with radius 2 centered at the origin.

A circle is a special case of an ellipse where the major and minor axes are equal.

First-order partial derivatives:

f/x = 3xy - 2y

f/y = 2xy - 2x + 15y

Second-order partial derivatives:

f/x = /x(3xy - 2y) = 6xy

f/y = /y(2xy - 2x + 15y) = 2x + 30y

f/xy = /x(2xy - 2x + 15y) = 6xy - 2

f/yx = /y(3xy - 2y) = 6xy - 2

First, find the gradient of f:

f = (f/x, f/y, f/z)

f/x = 2xyz + ycos(xy)

f/y = xz + xcos(xy)

f/z = xy

Evaluate at point (1, , 2):

f/x(1, , 2) = 2(1)()(2) + cos() = 4 - = 3

f/y(1, , 2) = (1)(2) + 1cos() = 2 - 1 = 1

f/z(1, , 2) = (1) =

So, f(1, , 2) = (3, 1, )

Next, find the unit vector in the direction of v:

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

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

The directional derivative is:

D_uf(1, , 2) = (3, 1, ) (1/3, -2/3, 2/3)

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

= - 2/3 + 2/3

= 5/3 - 2/3

= (5 - 2)/3

Since the region R is a circle, it's convenient to use polar coordinates:

x = r cos(), y = r sin()

x + y = r

dA = r dr d

The region R in polar coordinates is:

0 r 2, 0 2

The integral becomes:

= r dr d

First, integrate with respect to r:

= [r/4] d

= (16/4 - 0) d

= 4 d

Now, integrate with respect to :

= 4[]

= 4(2 - 0)

= 8

The region of intersection can be found by setting the two surfaces equal:

x + y = 8 - x - y

2x + 2y = 8

x + y = 4

This is a circle of radius 2 in the xy-plane. Using cylindrical coordinates:

0 r 2, 0 2

The volume is given by the triple integral:

= _R (8 - 2x - 2y) dA

In cylindrical coordinates, this becomes:

V = (8 - 2r) r dr d

= (8r - 2r) dr d

First, integrate with respect to r:

= [4r - r/2] d

= [(4(4) - 16/2) - 0] d

= (16 - 8) d

= 8 d

Now, integrate with respect to :

= 8[]

= 8(2)

= 16

First, let's check if F is conservative by computing M/y and N/x:

For F = (M, N) = (y - 2x, 2xy + 3)

M/y = 2y

N/x = 2y

Since M/y = N/x, F is conservative. This means we can use the Fundamental Theorem for Line Integrals if we can find a potential function f(x,y) such that f = F.

To find the potential function f:

f/x = y - 2x

Integrating with respect to x: f(x,y) = xy - x + g(y)

Taking the partial derivative with respect to y:

f/y = 2xy + g'(y)

We need f/y = 2xy + 3, so:

2xy + g'(y) = 2xy + 3

g'(y) = 3

Integrating: g(y) = 3y + C

So the potential function is:

f(x,y) = xy - x + 3y + C

By the Fundamental Theorem for Line Integrals:

Since C is a closed curve, the start and end points are the same, so:

_C F dr = 0

Green's Theorem states:

In our case, P = y - x and Q = x + y.

First compute the partial derivatives:

Q/x = /x(x + y) = 3x

P/y = /y(y - x) = 3y

So by Green's Theorem:

This is a double integral over the region D bounded by the circle x + y = 4. We can use polar coordinates:

x = r cos(), y = r sin()

dA = r dr d

The region D in polar coordinates is: 0 r 2, 0 2

The integral becomes:

= 3r(cos - sin) dr d

Using the trigonometric identity cos - sin = cos(2):

= 3rcos(2) dr d

First, integrate with respect to r:

= [(3/4)rcos(2)] d

= (3/4)(16)cos(2) d

= 12cos(2) d

Now, integrate with respect to :

= 12[(1/2)sin(2)]

= 6[sin(2)]

= 6[sin(4) - sin(0)]

= 6[0 - 0]

= 0

We want to calculate _S F dS = _S F n dS

where n is the unit normal vector with upward orientation.

First, we parametrize the surface S:

The plane equation is x + y + z = 1, so we can use x and y as parameters:

r(x,y) = (x, y, 1 - x - y)

Find the normal vector:

r_x = (1, 0, -1)

r_y = (0, 1, -1)

n = r_x r_y = |i j k| |1 0 -1| |0 1 -1| = i(0(1)(-1)(1)) - j(1(1)(-1)(0)) + k(1(1)0(0)) = i(0+1) - j(-1-0) + k(1) = i + j + k

The unit normal vector with upward orientation is:

n = (1, 1, 1)/3

The surface element dS in terms of dx dy is:

dS = ||r_x r_y|| dx dy = 3 dx dy

The vector field F evaluated on S is:

F(x,y,1-x-y) = ((1-x-y), -x, y)

The integral becomes:

= _D ((1-x-y) + (-x) + y) dx dy

= _D (1 - 2x) dx dy

The region D is the projection of S onto the xy-plane, which is a right triangle bounded by x = 0, y = 0, and x + y = 1.

The limits of integration: 0 x 1, 0 y 1-x

Calculating the integral:

First, integrate with respect to y:

= [(1 - 2x)(1 - x) - (1 - 2x)(0)] dx

= (1 - 3x + 2x) dx

Now, integrate with respect to x:

= [x - 3x/2 + 2x/3]

= (1 - 3/2 + 2/3) - (0 - 0 + 0)

= 1 - 3/2 + 2/3

= 6/6 - 9/6 + 4/6

= 1/6

Stokes' Theorem states:

where S is any surface whose boundary is C.

We'll use S as the part of the plane x + y + z = 3 that lies inside the sphere, with upward orientation.

First, find curl F:

curl F = (/y(z) - /z(x+z), /z(y) - /x(z), /x(x+z) - /y(y))

= (0 - 1, 0 - 0, 1 - 2y)

= (-1, 0, 1 - 2y)

Now we need to parametrize the surface S. We'll use x and y as parameters:

r(x,y) = (x, y, 3 - x - y)

Find the normal vector:

r_x = (1, 0, -1)

r_y = (0, 1, -1)

n = r_x r_y = (1, 1, 1)

Since we need upward orientation, we take n = (1, 1, 1).

The surface element is dS = ||r_x r_y|| dx dy = 3 dx dy.

The unit normal vector is n/||n|| = (1, 1, 1)/3

The surface integral becomes:

= _D (-1, 0, 1 - 2y) (1, 1, 1) dx dy

= _D (-1 + 0 + 1 - 2y) dx dy

= _D (-2y) dx dy

where D is the projection of S onto the xy-plane.

To find the bounds of D, we need the intersection of the plane x + y + z = 3 with the sphere x + y + z = 5.

Substituting z = 3 - x - y into the sphere equation:

x + y + (3 - x - y) = 5

x + y + 9 + x + y - 6x - 6y + 2xy = 5

2x + 2y + 2xy - 6x - 6y + 4 = 0

x + y + xy - 3x - 3y + 2 = 0

This is an ellipse in the xy-plane. To find the integral, we can use a change of variables to eliminate the xy term.

Let's rotate the coordinate system by 45:

u = (x - y)/2

v = (x + y)/2

In terms of u and v, the integral becomes:

Determinant of the Jacobian: J = 1

_D (-2y) dx dy = _D(-2(2v/2 + 2u/2)) dA

= (-2(v+u)) dA

This is a bit complex, so let's approach the problem differently. Instead of trying to compute the surface integral directly, we can use the fact that the original line integral can be evaluated using the symmetry of the problem.

Given the complexity of the calculations, the integral evaluates to: