SCalcET7 15.3.019.
Evaluate the double integral.
(No Response)
2
y
2
dA
,
D
is the triangular region with vertices (0, 1), (1, 2), (4, 1)
D
2
2
=
1

9.
–/1 points
SCalcET7 15.3.029.
Find the volume of the given solid.
Bounded by the cylinders
z
=
7
x
2
,
y
=
x
2
and the planes
z
= 0,
y
= 4
(No Response)
Solution or Explanation
Click to View Solution

13. 11. 9.
HW #19
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10.
–/1 points
SCalcET7 15.3.030.
Find the volume of the given solid.
Bounded by the cylinder
and the planes
in the
first octant
is odd with respect to
x
and
is odd with respect to
y
, and the region of integration is
symmetric with respect to both
x
and
y
, so
represents the volume of the solid region under the graph of
and above the rectangle
D
, namely a half circular cylinder with radius
a
and
length 2
b
(see the figure) whose volume is
Thus
y
2
+
z
2
=
16
x
= 2
y
,
x
= 0,
z
= 0
(No Response)
Solution or Explanation
Click to View Solution
11.
–/1 points
SCalcET7 15.3.067.
Use geometry or symmetry, or both, to evaluate the double integral.
(No Response)
Solution or Explanation
Now
is odd with respect to
x
and
is odd with respect to
y
, and the region of integration is
symmetric with respect to both
x
and
y
, so
represents the volume of the solid region under the graph of
and above the rectangle
D
, namely a half circular cylinder with radius
a
and
length 2
b
(see the figure) whose volume is
Thus
y
2
+
z
2
=
16
x
= 2
y
,
x
= 0,
z
= 0
(
4
ax
9
+
8
by
9
+
)
dA
,
D
= [−
a
,
a
]
×
[−
b
,
b
]
a
2
−
x
2
D
=
4
+
8
+
(
4
ax
9
+
8
by
9
+
)
dA
a
2
−
x
2
D
ax
9
dA
D
by
9
dA
D
dA
.
a
2
−
x
2
D
ax
9
by
9
4
=
8
ax
9
dA
D
by
9
dA
= 0.
D
dA
a
2
−
x
2
D
z
=
a
2
−
x
2
·
π
r
2
h
=
π
a
2
(2
b
) =
π
a
2
b
.
1
2
1
2
(
4
ax
9
+
8
by
9
+
)
dA
= 0 + 0 +
π
a
2
b
=
π
a
2
b
.
a
2
−
x
2
D

13. 11. 9.
HW #19
12.
–/1 points
SCalcET7 15.3.063.
Use geometry or symmetry, or both, to evaluate the double integral.
(No Response)
x
.
D

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