Solution to Problem Set #7
1. (a) (5 pt) If f (x, y) gives the pollution density, in micrograms per
square meter, and x and y are measured
in meters, give the units
RR
and practical interpretation of
f
(x,
y)
dA.
R
RR
micrograms
Solution. The unit of
f (x,R y)
R
R dA is meter2 · meter · meter =
micrograms. The meaning of
f (x, y) dA is the total amount of
R
pollution is the region A.
(b) (5 pt) Using Riemann sums with two subdivisions in each direction, find upper and lower bounds for the volume under the graph
of f (x, y) = 2 + xy above the rectangle R with 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.
Solution. Let f (x, y) = 2 + xy. We have fx = y ≥ 0 and fy = x ≥ 0 in
the rectangle R. The lower estimate is f (0, 0) · 1 + f (1, 0) · 1 + f (0, 1) ·
1 + f (1, 1) · 1 = 2 + 2 + 2 + 3 = 9.
The upper estimate is f (1, 1) · 1 + f (1, 2) · 1 + f (2, 1) · 1 + f (2, 2) · 1 =
3 + 4 + 4 + 6 = 17.
2. (12
p pt) Find the volume of the solid bounded by the surface z =
y y 2 + x and the planes x = 0, x = 1, y = 0, y = 1 and z = 0.
R1R1 p
Solution. The volume is 0 0 y y 2 + xdydx Let u = y 2 + x. Then du =
R p
R 1/2
2
3/2
3/2
2ydy, ydy = du
and y y 2 + xdy = u 2 du = u 3 + C = (y +x)
+ C.
2
3
R1R1 p
R 1 (y2 +x)3/2 1
So 0 0 y y 2 + xdydx = 0
dx
3
0
1
R 1 (1+x)3/2 (x)3/2
5/2
2(1+x)5/2
2(x)5/2 2
2
= 2(2)15 − 15
−
dx
=
−
− ( 15
− 0)
= 0
3
3
15
15
=
2(2)5/2
15
−
4
15
=
√
8 2
15
0
−
4
15
3. Compute the
iterated integrals.
R 1 following
R1
xy
√
(a) (12 pt) 0 0
dydx
2
2
x +y +1
Solution.
R1R1
0
0
√
xy
x2 +y 2 +1
dydx.
R
and √ 2xy 2 dy =
Let u = x2 + y 2 + 1. Then du = 2ydy, ydy = du
2
x +y +1
R x
R xu− 12
R1R1
1
1
2
2
√ du =
du = xu 2 = x(x + y + 1) 2 . Hence 0 0 √ 2xy 2 dydx =
2
2 u
x +y +1
R1
R
1 1
1
1
1
x(x2 + y 2 + 1) 2 dx = 0 (x(x2 + 2) 2 − x(x2 + 1) 2 )dx
0
0
3
3 1
3
3
2
2
(x2 +1) 2 (3) 2
(2) 2
= (use substitution u = x2 + 2)( (x +2)
−
)
=
(
−
)−
3
3
3
3
0
3
√
√
2
( (2)3 − 31 ) = 3 − 4 3 2 + 31 .
MATH 2850: page 1 of 3
Problem Set #7
MATH 2850: page 2 of 3
√
y
e
x + ey dxdy
0 0
R1R4 √
Solution. 0 0 ey x + ey dxdy.
R √
R √
Let u = x + ey . Then du = dx and ey x + ey dx = ey udu =
R y 1/2
y
y )3/2
y 3/2
e u du = 2e 3u = 2e (x+e
.
3
R 1 R 4 y√
R1 y
R 1 R 4 y√
y )3/2 4
Hence 0 0 e x + ey dxdy = 0 0 e x + ey dxdy = 0 2e (x+e
dy
3
0
5y 1
R 1 2ey (4+ey )3/2 2ey (ey )3/2
R 1 2ey (4+ey )3/2 2e 5y2
4(4+ey )5/2
4e 2 −
dy = 0
− 3 dy =
− 15 =
3
3
3
15
0
(b) (12 pt)
R1R4
5/2
( 4(4+e)
15
5
−
4e 2
15
)−
RR
5/2
( 4(5)15
−
4
)
15
=
4(4+e)5/2
15
0
5
−
4e 2
15
−
4(5)5/2
15
+
4
.
15
4y
dA
D x3 +4
where D = {(x, y)|1 ≤ x ≤ 6, 0 ≤ y ≤ 4x}.
R R 4y
R 6 R 4x 4y
R 6 2y2 4x
R 6 32x2
Solution.
dA
=
dydx
=
dx. Let
dx
=
3
3
3
D x +4
1 0 x +4
1 x +4
1 x3 +4
0R
R
2
32
u = x3 + 4. Then du = 3x2 dx, x2 dx = u3 du and x32x
du =
3 +4 dx =
3u R
R
6
4y
32
32
32
3
3
ln
|u|
+
C
=
ln
|x
+
4|
+
C.
Hence
dA
=
ln
|x
+
4|
=
3
3
3
D x3 +4
4. (12 pt) Evaluate
32
3
ln(220) −
32
3
1
ln(5).
RR p
5. (12 pt)Evaluate
x y 2 − x2 dA where D = {(x, y)|0 ≤ x ≤ y, 0 ≤ y ≤ 3}
D
RR p
R R p
R R
1
2 − x2 dA = 3 y x
2 − x2 dxdy = 3 y x(y 2 −x2 ) 2 dxdy.
Solution.
x
y
y
D
0 0
0 0
1
R
2
Let u = y 2 − x2 . Then du = −2xdx, xdx = − u2 du and − (u)2 du =
R3
R 1 3
R3Ry
1
y
− 3 u 2 = − 31 (y 2 −x2 ) 32 . Hence 0 0 x(y 2 −x2 ) 2 dxdy = 0 − 13 (y 2 −x2 ) 23 dy =
0
3
R3 1 3
R3 1 2 3
1
81
27
4
(y ) 2 dy = 0 3 y dy = 12 y = 12 = 4 .
0 3
0
6. (15 pts)
R1R1
0
y
2
ex dxdy
Solution. Let D = {(x, y)|y ≤ x ≤ 1, 0 ≤ y ≤ 1} Then 0 ≤ y ≤ x and
0 ≤ x ≤ 1. So D is the same as {(x, y)|0 ≤ x ≤ 1,
0 ≤ y ≤ x}.
R 1 R 1 x2
R 1 R x x2
R 1 x2 x
R1 2
x2
We have 0 y e dxdy = 0 0 e dydx = 0 e y dx = 0 ex xdx = e2 |10 =
e
2
0
− 21 .
7. (15 pts)
R1R1
0
x2
x3 sin(3y 3 )dydx
Solution. Let D = {(x, y)|0 ≤ x ≤ 1, x2 ≤ y ≤ 1}. Since x2 ≤ y, we
√
√
have x ≤ y, 0 ≤ x ≤ y and 0 ≤ y ≤ 1. So D is the same as
√
{(x, y)|0 ≤ x ≤ y, 0 ≤ y ≤ 1}.
Problem Set #7
MATH 2850: page 3 of 3
R1R1
3
3
R 1 R √y
3
3
We have 0 x2 x sin(3y )dydx = 0 0 x sin(3y )dxdy =
R 1 y2
3)
1
sin(3y 3 )dx = − cos(3y
|10 = − cos(3)
+ 36
.
36
36
0 4
R1
x4
0 4
√y
sin(3y ) dx =
3
0