Download Calculus III Exam 3 Solutions: Polar, Cylindrical, Vector Calculus & Green's Theorem - Pro and more Exams Advanced Calculus in PDF only on Docsity!
Math 2433 Calculus III —– Exam 3
University of Houston
May 1, 2012
Name (printed):
Last four digits of your PeopleSoft ID:
Your recitation class (please circle one):
Zhiheng Liu (9:00-10:00 am, SEC 206)
Nidhi Jaiswal (9:00-10:00 am, SEC 205)
Ricky Ng (8:00-9:00 am, SEC 102)
Ilija Jegdic (9:00-10:00 am, SEC 204)
Closed books and notes. NO CALCULATORS. Show all your work for all credits. An-
swers, even correct obtained through guesswork, incorrect procedure or no procedure, are
worthless.
- (13 points) Using polar coordinates, evaluate the integral
D
sin(x
2
2
) dx dy
where D is the region 1 ≤ x
2
2 ≤ 4.
Solution: Using the polar coordinates, write
(r, θ)
0 ≤ θ ≤ 2 π
f ix θ, 1 ≤ r ≤ 2
Then
Ω
sin(x
2
2
) dx dy =
2 π
0
2
1
sin(r
2
)r dr dθ = ...
- (12 points) Let f (x, y) = (x + y, 2 xy) be a vector-valued function. Calculate its
Jacobian determinant det D(f ).
Solution: f = (f 1
, f 2
) where f 1
= x + y and f 2
= 2xy. Then
det D(f ) = det
∂f 1
∂x
∂f 2
∂x
∂f 1
∂y
∂f 2
∂y
= det
1 2 y
1 2 x
= 2x − 2 y.
- (13 points) Let C be the curve given by
~r(t) = (1 + sin t)
i + (1 + sin
2
t)
j, 0 ≤ t ≤
π
and
F is the vector field
F (x, y) = 2x ~i + 2y ~j.
Verify that
F = ▽ f where f (x, y) = x
2
2 , and then compute the work
C
F (~r) · d~r done
by
F on a particle moving along C.
Solution: ▽ f = ▽(x
2
2 ) = 2x~i + 2y~j =
F. Then by the fundamental theorem of
calculus, the work
W =
C
F (~r) · d~r = f (~r(
2 i
)) − f (~r(0)) = f (1 + 1, 1 + 1) − f (1, 1) = ...
- (12 points) Let ~v = xy~i − 2 yz~j. Calculate the divergence div ~v and the curl curl ~v.
Solution: v 1
= xy, v 2
= − 2 yz, v 3
= 0. Then
div ~v =
∂v 1
∂x
∂v 2
∂y
∂v 3
∂z
= y − 2 z,
and
curl ~v = det
i
j
k
∂
∂x
∂
∂y
∂
∂z
v 1
v 2
v 3
= det
i
j
k
∂
∂x
∂
∂y
∂
∂z
xy − 2 yz 0
= det
∂
∂y
∂
∂z
− 2 yz 0
i − det
∂
∂x
∂
∂z
xy 0
j + det
∂
∂x
∂
∂y
xy − 2 yz
k
= (0 + 2y)
i − (0 − 0)
j + (0 − x)
k = 2y~i − x
k.
- (13 points) By applying Green’s theorem to evaluate the line integral
C
x
2
y dx +
x
3
dy
where C is the square Ω with vertices (0, 0), (2, 0), (2, 2) and (0, 2).
Solution: Let P = x
2 y and Q =
1
3
x
3
. Applying Green theorem, we have
C
x
2 y dx+
1
3
x
3 dy =
∫ ∫
Ω
∂Q
∂x
∂P
∂y
)dxdy =
Ω
(x
2 − x
2 )dxdy = 0.
- (12 points) Find the area of the ellipse
x
2
4
y
2
9
= 1 by using the formula area =
1
2
C
x dy − y dx.
Solution: The area is 6π (see an example in the lecture notes, page 205).
