1. Let F(x, y) =
y cos(xy)
x cos(xy)
, and let γ(t) be the path
√
R
t
√
for t ∈ [0, π 2 ]. Find γ F · ds.
sin( t)
R
Rb
Solution: Remember that the standard formula for γ F · ds, for t ∈ [a, b], is a (F ◦ γ)(t) · γ 0 (t)dt. In
!
√
√
√
1
√
sin(
t) cos(
t sin(
t))
2 t√
0
√
√
√
this case, γ (t) =
; (F ◦ γ)(t) =
. Shoving all of this into the
1
√
cos( t)
t cos( t sin( t))
2 t
integral, and using a = 0 and b = π 2 , we’re looking at
Z
0
π2
√
√
√
√
√
√
√
1
1
(sin( t) cos( t sin( t)) √ + t cos( t sin( t)) √ cos( t))dt
2 t
2 t
Your reaction to this shouldn’t be “How do I do this integral?”. It should be “How can I avoid doing
this integral?” There’s only one tool we have for doing a line integral of a vector field without actually doing
the integral - namely, if F happens to be conservative, then all we have to do is find a potential function f
and find f (γ(π 2 )) − f (γ(0)). Even better, F is conservative if and only if a potential function exists - so we
might as well just go ahead and try to find one.
y cos(xy)
f is a potential function for F if and only if F = 5f . So we’re looking for an f so that
=
x cos(xy)
fx
. fx = y cos(xy). Thinking of y as a constant, this is just saying “the derivative of f is y cos(xy)”;
fy
we know how to solve that for f . That’s what integrals are for. So integrating both sides - remember, we’re
thinking of y as a constant, so we’re integrating dx - we get f = sin(xy) + c. Importantly, c is only a constant
as far as x is concerned, so c might depend on y. In other words, we have f (x, y) = sin(xy) + g(y) for some
mystery function g. Now we need to find out what g is.
We have one more piece of information: fy = x cos(xy). But we know that f = sin(xy) + g(y); taking
the y-partial of both sides, we get fy = x cos(xy) + g 0 (y). Plugging this into the equation fy = x cos(xy),
we get x cos(xy) + g 0 (y) = x cos(xy). Cancelling, g 0 (y) = 0. Again, this is a statement about the derivative
of a function; we can solve it with an integral. Integrating both sides (dy, since g only knows about y) we
get g(y) = c, where c is some constant. So f (x, y) = sin(xy) + g(y) = sin(xy) + c. But we only care about
getting a potential function; we don’t need all of them. So we can just pick whatever value of c we like best.
I’ll pick 0. So our potential function is f (x, y) = sin(xy).
2
R f (γ(0)) = f (0, sin(0)) = f (0, 0) = sin(0 · 0) = 0. f (γ(π )) = f (π, sin(π)) = f (π, 0) = sin(π · 0) = 0. So
F · ds = 0 − 0 = 0.
γ
2. For each of the following vector fields F, determine whether or not F is conservative. If it is, find a
potential function for it.
3x
a. F(x, y) =
2y
2y
b. F(x, y) =
3x
2
x −y
c. F(x, y) =
y2 − x
1

x3
d. F(x, y, z) =  3z 2 
2y
sin(y) + x
e. F(x, y) =
x cos(y)

Solution: First, let’s think about strategy. There’s an easy way to determine whether something’s
conservative, without going through the hassle of trying to find a potential
function.
The idea is this: If F
∂f /∂x
is conservative, then it’s the gradient of some function f . So it’s
(assuming we’re doing this
∂f /∂y
∂ ∂f
∂ ∂f
in two dimensions). But remember that mixed partials are equal ; that is, fxy = fyx . So ∂y
∂x = ∂x ∂y . In
other words, if we take the y-partial of the x-component of our vector field, we should get the same thing as
if we took the x-partial of the y-component. If that doesn’t happen, then there couldn’t have been any such
f.
If that explanation didn’t make a whole lot of sense, hang on. We’ll do a few examples below, which will
hopefully clear things up.
3x
∂
3x = 0. Taking the
. Taking the y-derivative of the x-component, we get ∂y
2y
∂
x-derivative of the y-component, we get ∂x
2y = 0. 0 = 0, so this vector field is conservative! That means
we need to look for a potential f . fx = 3x; integrating both sides dx, we get f = 23 x2 + c; but since we were
integrating dx, c is allowed to depend on y. So we really have f (x, y) = 32 x2 + g(y). Taking the y-partial
of both sides of this, we have fy = g 0 (y). But fy = 2y if F is going to be the gradient of f ; so g 0 (y) = 2y.
Therefore g(y) = y 2 + c. So f (x, y) = 32 x2 + y 2 + c. But we were only asked for a potential function, so we
might as well take c = 0, giving us f (x, y) = 23 x2 + y 2 .
(a): F(x, y) =
2y
∂
. Taking the y-derivative of the x-component, we get ∂y
2y = 2. Taking the
3x
∂
3x = 3. 2 6= 3, so if F were the gradient of some function f , then
x-derivative of the y-component, we get ∂x
the mixed partials of f wouldn’t match, which is impossible. So F isn’t conservative, and we’re done.
(b): F(x, y) =
x2 − y
∂
∂
(c): F(x, y) =
. ∂y
(x2 −y) = −1, ∂x
(y 2 −x) = −1, and −1 = −1; so F is conservative. So we
y2 − x
need to look for a potential function f . fx = x2 −y; integrating both sides dx, we get f (x, y) = 31 x3 −xy+g(y).
Taking the y-partial of both sides of this, we get fy = −x + g 0 (y). But fy = y 2 − x, because F is supposed
to be the gradient of f ; so y 2 − x = −x + g 0 (y). So g 0 (y) = y 2 , and therefore g(y) = 31 y 3 + c. So f (x, y) can
be 31 x3 − xy + 13 y 3 + c for any c; so f (x, y) = 13 x3 − xy + 13 y 3 works.

x3
(d): F(x, y, z) =  3z 2 . This one’s a little different - it’s three-dimensional, instead of two. Our
2y
mixed-partial trick can still work, but we would have to be careful - there’s no reason, for example, why the
y-partial of the x component and the x-partial of the z component would match. Instead of trying to hash
out exactly how we need to adjust the trick, let’s just go ahead and try to find a potential function. If we
can’t - if we actually run into a dead end, I mean - then we know it wasn’t conservative, as long as we didn’t
make any choices along the way.
If f will be the potential function, then fx = x3 . Integrating dx, we have f (x, y, z) = 41 x4 + c. But
now we have two variables that are constant as far as x is concerned, so c might depend on either one! So

2
f (x, y, z) = 14 x4 + g(y, z) for some mystery function g.
We also need fy = 3z 2 . Since f = 14 x4 + g(y, z), taking the y-partial of both sides we get fy = gy (y, z).
So gy (y, z) = 3z 2 . Integrating both sides of this dy, we get g(y, z) = 3yz 2 + c - except that now c is allowed
to depend on the one remaining variable z, so g(y, z) = 3yz 2 + h(z). Plugging this into our equation for f ,
we have f (x, y, z) = 41 x4 + 3yz 2 + h(z).
Finally, we need fz = 2y. So 6yz + h0 (z) = 2y. Then h0 (z) = 2y − 6yz. But - oops! h0 (z) only knows
about z. It’s never heard of y. So the fact that y is still there (and doesn’t cancel out or anything) means
we’ve hit an impossible situation. So no such h exists, so no such f exists. F wasn’t conservative.
sin(y) + x
∂
∂
(sin(y) + x) = cos(y), ∂x
(x cos(y)) = cos(y), and cos(y) = cos(y), so
(e): F(x, y) =
. ∂y
x cos(y)
this vector field is conservative. That means we need to look for a potential function f .
fx = sin(y) + x, so f = x sin(y) + 12 x2 + g(y). Then fy = x cos(y) + g 0 (y). But fy = x cos(y), so
x cos(y) = x cos(y) + g 0 (y). g 0 (y) = 0, so g(y) = c for some constant c; we might as well pick 0. So
f (x, y) = x sin(y) + 21 x2 is a potential function.
3. Let P be the plane given by 2x + 3y + 4z = 5. (a) Find a parameterization for P , and then (b) find
the surface area of the part of P with 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1.
Solution: To parameterize P , we get to use two parameters, s and t. We want a function f (s, t) that
outputs three-dimensional points
 onthis plane. To make it easy on ourselves, let’s just start by trying
s
x = s, y = t (so f (s, t) =  t ). If x = s, y = t, and (x, y, z) is on the plane, then it must be
???
that 2s + 
3t + 4z = 5. So4z = 5 − 2s − 3t, and therefore z = 45 − 12 s − 34 t. So our parameterization is
s
. (Note: There are lots of different possible parameterizations. Personally, I
t
f (s, t) = 
1
3
5
4 − 2s − 4t
think this one is the most intuitive, but feel free to find your own.)
R Now we need the surface area. Remember the formula for the surface area of a parameterized surface:
||fs × ft ||dA. We need to know three things: fs , ft , and D. Let’s start with D.
D
D is the domain of our parameterization; in other words, it’s the region of the s − t plane so that f (s, t)
is in the part of the plane we care about. That’s where x and y are both between 0 and 1. But in our
parameterization, x = s and y = t; so that’s exactly where s and t are both between 0 and 1. So D is just
the square [0, 1] × [0, 1].




∂
1
∂s s
∂
 =  0 . ft is the t-partial of
fs is the s-partial of f , which means fs = 
∂s t
1
3
∂ 5
−1/2
( − s − 4 t)

 
 ∂s 4 2


∂
s
0
1/2
q
∂t
∂
 =  1 . So fs × ft =  3/4 . So ||fs × ft || = 1 + 9 + 1 =
f , so ft = 
t
4
16
∂t
∂ 5
−3/4
−1
( 4 − 12 s − 34 t)
∂t
√
p
29/16 = 429 .
√
R
R1R1 √
Finally, then, D ||fs × ft ||dA = 0 0 429 dsdt = 429 (I’ll let you check that yourself, for some doubleintegral practice). And now we’re done - that’s the area.
3
4. f (s, t) = hs2 , st, t2 i, for s and t between 0 and 1. Compute the surface area of the surface defined by
f.
R
Solution: Again, we’ll use the formula D ||fs × ft ||dA. Again, s and t both range freely from 0 to 1, so
R1R1
D is justthe unit
integral can bewritten 0 0 ||fs × ft ||dsdt.
 square;
 our
2s
0
2t2
√
√





t
s
−4st . ||fs ×ft || = 4t4 + 16s2 t2 + 4s4 = 2 t4 + 4s2 t2 + s4 .
fs =
, ft =
. So fs ×ft =
0
2t
2s2
R1R1 √
So we’re looking at 0 0 2 t4 + 4s2 t2 + s4 dsdt.
This integral doesn’t actually have a nice solution (go ahead, try plugging it into Wolfram Alpha, see
what you get) so I’ll leave it there.
4