Download Fourth Practice Test, Math 234 (1) Find the volume of the solid

Fourth Practice Test, Math 234
(1) Find the volume of the solid bounded above by the paraboloid z = 9 − x2 −
y 2 , below by the xy–plane, and lying outside the cylinder x2 + y 2 = 1.
(2) Find the area of the surface 2x3/2 + 2y 3/2 − 3z = 0 above the rectangle
R : 0 ≤ x , y ≤ 1 in the xy–plane.
(3) Compute the line integral of the function f (x, y) = x + y over the curve
C : x2 + y 2 = 4
in the first quadrant from (2, 0) to (0, 2).
(4) Find the work done by the force
F = (xy, y − x)
over the straight line from (1, 1) to (2, 3).
(5) Use Green’s theorem to evaluate the following integral
I
(6y + x)dx + (y + 2x)dy
C
where C is the circle (x−2)2 +(y−3)2 = 4 with counterclockwise orientation.
1
2
Solutions
(1) The most convenient coordinates for this problem are cylindrical coordinates. We set up a triple integral which computes the volume. Starting
with the integration in the z–direction, it ranges from z = 0 (xy–plane) to
z = 9 − r2 , the paraboloid. The integration in the r–direction ranges from
r = 1 (the cylinder x2 + y 2 = 1) to r = 3 (that is where the paraboloid
intersects the xy–plane). The θ–integration ranges from 0 to 2π. Hence the
volume is given by
Z 2π Z 3 Z 9−r2
Z 2π Z 3
r dz dr dθ =
r(9 − r2 ) dr dθ
0
1
0
0
1
r=3
9 2 1 4 r − r 2
4
r=1
81 17
−
= 2π
4
4
= 32π.
=
2π
(2) The surface is given as a level set {f (x, y, z) = 0} for the function f (x, y, z) =
2x3/2 + 2y 3/2 − 3z. We choose p = (0, 0, 1), and we compute
√
√
∇f (x, y, z) = (3 x, 3 y, −3),
p
p
|∇f (x, y, z)| = 9x + 9y + 9 = 3 x + y + 1
|∇f (x, y, z) • p| = 3
so that the surface area is given by
Z Z p
Z 1Z 1p
x + y + 1 dx dy =
x + y + 1 dx dy
R
0
=
2
3
Z
0
1
(x + y + 1)3/2 |x=1
x=0 dy
0
Z
Z
2 1
2 1
3/2
=
(y + 2) dy −
(y + 1)3/2 dy
3 0
3 0
4
4
=
(y + 2)5/2 |10 − (y + 1)5/2 |10
15
15
√
√
4
(1 − 8 2 + 9 3).
=
15
(3) The curve C is the circle centered at the origin with radius 2 which we may
parameterize by
r(t) = (2 cos t, 2 sin t) , 0 ≤ t ≤ 2π.
so that
|v(t)| = 2
and
Z
Z
f ds =
4(cos t + sin t) dt
C
0
=
π/2
8
3
(4) We write the straight line as
r(t) = (1 − t)(1, 1) + t(2, 3) = (1 + t, 1 + 2t) , 0 ≤ t ≤ 1
so that
v(t) = (1, 2)
and the work done is given by
Z 1
Z 1
((1 + t)(1 + 2t), 1 + 2t − (1 + t)) • (1, 2) dt
F(r(t)) • v(t) dt =
0
0
Z
1
(2t2 + 3t + 1, t) • (1, 2) dt
=
0
Z
=
1
(2t2 + 5t + 1) dt
0
1
2 3 5 2
=
t + t + t
3
2
0
25
.
=
6
(5) The integral is of the form
I
Z Z ∂N
∂M
M dy − N dx =
+
dx dy
∂x
∂y
C
R
if we choose
M = y + 2x , N = −6y − x
so that
∂M
∂N
=2,
= −6
∂x
∂y
Hence we have to compute
Z Z
−4
dx dy
R
which is −4 times the area of R. Since R is a circle with radius 2, the final
answer is
−16π.