c Dr Oksana Shatalov, Spring 2014
1
Spring 2014 Math 251
Week in Review 6
courtesy: Oksana Shatalov
(covering Sections 13.1-13.3 )
13.1-13.3 : Double Integral
Key Points
• FUBINI’s THEOREM:If f is continuous on on the rectangle R = [a, b] × [c, d] then
b
Z
ZZ
d
Z
Z
d
Z
a
R
c
c
b
f (x, y) dxdy.
f (x, y) dydx =
f (x, y) dA =
a
• If g and h are continuous functions of one variable and R = [a, b] × [c, d] then
! Z
!
ZZ
Z
b
g(x)h(y) dA =
d
g(x)dx
R
h(y) dy .
a
c
• If f (x, y) ≥ 0 and f is continuous on the region D then the volume V of the solid S that lies above D and
under the graph of f , i.e. S = (x, y, z) ∈ R3 | 0 ≤ z ≤ f (x, y), (x, y) ∈ D , is
ZZ
V =
f (x, y) dA.
D
• If D is a region of type I such that D = {(x, y)| a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x)} then
ZZ
b
Z
g2 (x)
Z
f (x, y) dA =
f (x, y) dydx.
D
a
g1 (x)
• If D is a region of type II s.t. D = {(x, y)| c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y)} then
ZZ
Z
d
Z
h2 (y)
f (x, y) dA =
D
f (x, y) dxdy.
c
h1 (y)
1. Reverse the order of integratioin the following integrals:
Z 1 Z 1−x2
(a)
f (x, y) dydx
√
− 1−x2
0
Z
3Z
9
(b)
f (x, y) dxdy
y2
0
Z
−2
Z
(c)
x+6
Z
0
x2
Z
f (x, y) dydx +
−6
0
f (x, y) dydx.
−2
0
2. Evaluate the integral
Z
0
3
Z
9
y2
y cos(x2 )dxdy.
c Dr Oksana Shatalov, Spring 2014
2
3. Find the volume of the solid that lies above the region D = {(x, y) : x2 + y 2 ≤ 2x, y ≥ 0}
and under the graph of ρ(x, y) = y.
ZZ
y cos y
dA , where R = {(x, y)| 1 ≤ x ≤ e4 , 0 ≤ y ≤ π/2}.
4. Find the integral
x
R
ZZ
√
5. Evaluate
D
y
dA where D is the region in the first quadrant bounded by x =
1 + x2
y 2 , x = 4, y = 0.
Z 1Z 2
x3 sin y 3 dy dx .
6. Evaluate
0
2x2
2
7. Sketch the region bounded by y 2 = 2x (or x = y2 ), the line x + y = 4 and the x-axis, in the
first quadrant. Find the area of the region using a double integral.
Z 2Z 4
(x2 + y 2 )dx dy and find the
8. Describe the solid which volume is given by the integral
0
volume.
9. Find the volume of the solid bounded by the following planes
z = 1 + x + y, z = 0, x + y = 1, x = 0, y = 0.
y2