Math 253 Homework assignment 6
R
1. Consider the integral R xy 2 dA, where A is the rectangle [0, 1] × [0, 1].
(a) Calculate the Riemann sum corresponding to this integral, with the subdivision
corresponding to ∆x = ∆y = 0.2 and using the centre of each small rectangle as
the sample point (x∗ij , yij∗ ).
(b) Using an iterated integral, calculate the value exactly.
2. Find the volume of the solid bounded by the planes x = 1, x = 2, y = 0, y = π/2,
z = 0 and the surface z = x cos y.
ZZ
√
x + ydA, where R = [0, 1] × [0, 3].
3. Calculate
R
ZZ
4. Find
(x2 + y 2 )dA, where R is the rectangle 0 ≤ x ≤ a,
0 ≤ y ≤ b.
R
π
Z
Z
x
5. Calculate the iterated integral
cos ydydx.
0
−x
1
and above the region in the xy-plane
x+y
bounded by x = 1, x = 2, y = 0 and y = x.
6. Find the volume under the surface z =
7. Using a double integral, calculate the volume of the tetrahedron in the first quadrant
bounded by the coordinate planes and the plane which intersects the x- y- and z-axes
at a, b and c, respectively, where a, b, c are positive numbers.
Z 1Z 1
3
8. Calculate the integral I =
ey dydx.
√
0
x
1
Math 253 Homework assignment 6
R
1. Consider the integral R xy 2 dA, where A is the rectangle [0, 1] × [0, 1].
(a) Calculate the Riemann sum corresponding to this integral, with the subdivision
corresponding to ∆x = ∆y = 0.2 and using the centre of each small rectangle as
the sample point (x∗ij , yij∗ ).
Solution: The Riemann sum has 25 terms: ((.1)(.1)2 + (.1)(.3)2 + (.1)(.5)2 +
(.1)(.7)2 + (.1)(.9)2 + (.3)(.1)2 + (.3)(.3)2 + . . . )(.2)(.2) = 0.16500 .
(b) Using an iterated integral, calculate the value exactly.
Z 1Z 1
Z 1
Z
3 1
1 1
2
Solution:
xy /3 y=0 dx =
xy dydx =
xdx =
3 0
0
0
0
1
1 2 1
=
x /2 0 =
= 0.16666 . . .
3
6
2. Find the volume of the solid bounded by the planes x = 1, x = 2, y = 0, y = π/2,
z = 0 and the surface z = x cos y.
2
Z π/2 Z 2
Z
Z π/2 2
x cos y
3 π/2
Solution: Vol =
x cos ydxdy =
dy =
cos ydy =
2
2 0
0
1
0
x=1
3
3
=
[sin y]π/2
0
2
2
ZZ
√
3. Calculate
x + ydA, where R = [0, 1] × [0, 3].
R
1
2
3/2
(x + y)
Solution:
x + ydA =
x + ydxdy =
dy =
3
0
0
0
R
x=0
Z
3
2 3
4 4 5/2
=
(1 + y)5/2 − y 5/2 0 =
[(4 − 35/2 ) − 1] =
((1 + y)3/2 − y 3/2 )dy =
3 0
15
15
√
4
[31 − 9 3]
15
ZZ
4. Find
(x2 + y 2)dA, where R is the rectangle 0 ≤ x ≤ a, 0 ≤ y ≤ b.
ZZ
R
Z 3Z
√
ZZ
Z
2
Solution:
a
1
Z
2
Z
a
0
a
y3
x y+
3
b
2
(x + y )dydx =
0
a
2
3
bx
a3 b + ab3
b3 x
b3
=
+
(bx + )dx =
=
3
3
3 0
3
0
Z πZ x
5. Calculate the iterated integral
cos ydydx.
0
−x
Z πZ x
Z π
Z
x
Solution:
cos ydydx =
[sin y]y=−x = 2
Z
3
b
2
(x + y )dA =
R
Z
√
0
dx =
y=0
2
0
−x
0
0
π
sin xdx = −2 [cos x]π0 = 4 .
1
and above the region in the xy-plane
x+y
bounded by x = 1, x = 2, y = 0 and y = x.
6. Find the volume under the surface z =
1
Z
Solution: Vol =
Z 2
ln(2)dx = ln 2
2
1
Z
0
x
1
dydx =
x+y
Z
Z
2
[ln(x +
y=x
y)]y=0
2
(ln(2x) − ln(x))dx =
=
1
1
1
7. Using a double integral, calculate the volume of the tetrahedron in the first quadrant
bounded by the coordinate planes and the plane which intersects the x- y- and z-axes
at a, b and c, respectively, where a, b, c are positive numbers.
x y
z
Solution: The plane has equation + + = 1, and it intersects the xy-plane in
a
b
c
x y
the triangle bounded by the axes and the line + = 1, or y = b(1 − ab x). So we may
a b
compute the volume as
b(1− xa )
Z a
Z a Z b(1− x ) a
x y
xy y 2
dydx = c
−
c 1− −
dx =
y−
a b
a
2b y=0
0
0
0
Z a
Z a
1 x
x x
x
1
x 2
x2
bc
1 − − (1 − ) − (1 − ) dx = bc
dx =
− +
a a
a
2
a
2 a 2a2
0
0
a
x x2
abc
x3
bc
−
+ 2 =
2 2a 6a 0
6
Z 1Z 1
3
8. Calculate the integral I =
ey dydx.
√
0
x
3
Solution: There is no nice expression for the antiderivative of ey , which by convention
3
means e(y ) , so we solve this problem by reversing the order of integration. Notice that
2
the
√ region of integration is the set in R defined by the inequalities 0 ≤ x ≤ 1 and
x ≤ y ≤ 1, or in other words, the region bounded by the y-axis, the line y = 1
and the curve x = y 2 . Thus we can calculate the double integral with the order of
integration reversed:
" 3 #1
Z 1 Z y2
Z 1
e−1
ey
3
3
I=
ey dxdy =
y 2 ey dy =
=
3
3
0
0
0
0
2