SAMPLE TEST TWO
(1) (a) Estimate the volume under the solid bounded by the xy-plane, the graph of
f (x, y) = x2 + 4y and above the rectangle R = [0, 2] × [0, 3] by using a double
Riemann sum with a grid of four rectangles bounded by the lines x = 1 and
y = 2. Evaluate the function at the upper right corner.
i
j ∆xi
∆yj
f (xi , yj )
∆xi ∆yj f (xi , yj )
1 1
1
2
f (1, 2) = 9
18
1 2
1
1
f (1, 3) = 13
13
2 1
1
2
f (2, 2) = 12
24
2 2
1
1
f (2, 3) = 16
16
Sum=
(b) Find the exact volume. Ans:
71
R2R3
0
0
x2 + 4y dy dx = 44
(2) Consider the solid bounded by the xy-plane and the two graphs: x2 + y 2 + z 2 = 4
and the cylinder x2 + y 2 = 1.
(a) Setup but do not evaluate an iterated integral for the volume using rectangular
R 1 R √1−x2 R √4−x2 −y2
coordinates. Ans: 0 −√1−x2 0
1 dz dy dx
(b) Setup the equivalent integral using polar coordinates and evaluate the inte√
R 2π R 1 R √4−r2
gral. Ans: 0 0 0
1r dz dr dθ = 2(8/3 − 3)π
(3) Find the volume of the region in the first octant bounded by the coordinate planes
R 2 R 3−3x/2 R 1−y/3−x/2
and the plane 3x + 2y + 6z = 6. Ans: 0 0
1 dz dy dx = 1
0
1
2
SAMPLE TEST TWO
(4) Use a double integral to find the volume of the solid under the paraboloid z = x2 +y 2
R 2π R 3 R r2
and above the disk x2 + y 2 ≤ 9. Ans: 0 0 0 1r dz dr dθ = 81π/2
(5) Use spherical coordinates to evaluate
Z √
Z Z √
16−x2
4
32−x2 −y 2
√
0
0
2π
Z
Z
√
4 2 Z π/4
Ans :
0
0
p
x2 + y 2 + z 2 dz dy dx.
x2 +y 2
√
ρρ2 sin(φ) dφ dρ, dθ = (256 − 128 2)(2π).
0
R1R1
R1Ry
(6) Evaluate the iterated integral: 0 x sin(y 2 ) dy dx Answer: Reverse order to 0 0 sin(y 2 ) dx dy =
R1
2
2
1
0 y sin(y ) dy = −cos(y )/2|0 = − cos(1)/2 + 1/2
(7) Find the double integral of f (x, y) = xy over the triangular region with vertices
R 1 R 1−x
(1, 0), (0, 1), (0, −1). Ans: 0 −1+x xy dy dx = 0 (Note function is odd in y)
(8) Interchange the order of integration in the following integral: You do not have to
evaluate the integral.
Z
1 Z x2
F (x, y) dy dx
0
Ans:
R1R1
0
√
y
0
F (x, y) dx dy
(9) Setup but you do not have to evaluate the integral for the area enclosed by one loop
R π/4 R cos(2θ)
of the four-leaved rose r = cos(2θ). Ans: −π/4 0
r dr θ