Respuesta :
he volume of the solid under a surface
z
=
f
(
x
,
y
)
and above a region D is given by the formula
∫
∫
D
f
(
x
,
y
)
d
A
.
Here
f
(
x
,
y
)
=
6
x
y
. The inequalities that define the region D can be found by making a sketch of the triangle that lies in the
x
y
−
plane. The bounding equations of the triangle are found using the point-slope formula as
x
=
1
,
y
=
1
and
y
=
−
x
3
+
7
3
.
Here is a sketch of the triangle:
Intersecting Region
The inequalities that describe D are given by the sketch as:
1
≤
x
≤
4
and
1
≤
y
≤
−
x
3
+
7
3
.
Therefore, volume is
V
=
∫
4
1
∫
−
x
3
+
7
3
1
6
x
y
d
y
d
x
=
∫
4
1
6
x
[
y
2
2
]
−
x
3
+
7
3
1
d
x
=
3
∫
4
1
x
[
y
2
]
−
x
3
+
7
3
1
d
x
=
3
∫
4
1
x
[
49
9
−
14
x
9
+
x
2
9
−
1
]
d
x
=
3
∫
4
1
40
x
9
−
14
x
2
9
+
x
3
9
d
x
=
3
[
40
x
2
18
−
14
x
3
27
+
x
4
36
]
4
1
=
3
[
(
640
18
−
896
27
+
256
36
)
−
(
40
18
−
14
27
+
1
36
)
]
=
23.25
.
Volume is
23.25
.
Following are the steps to calculate volume:
In this question the slope point form:
[tex]\bold{y-y_1=m(x-x_1)}\\\\\bold{y-2=\frac{1}{3}(x-1) \longrightarrow \text{for point}\ (1,2)}\\\\\bold{y=\frac{-x}{3}+\frac{7}{3}}[/tex]
Volume calculation:
[tex]\to \int^{4}_{1} \int^{\frac{-x}{3}+\frac{7}{3}}_{1} \ 6xy \ dy\ dx\\\\\to \int^{4}_{1} [ 3x^2y]^{\frac{-x}{3}+\frac{7}{3}}_{1}\ dx\\\\\to \int^{4}_{1} ( \frac{(x-7)^2 x}{3} -3x)\ dx\\\\\to \frac{256}{9} -\frac{187}{86}\\\\\to \frac{93}{4}[/tex]
Therefore the final answer is "[tex]\bold{\frac{93}{4}}[/tex]".
Learn more:
brainly.com/question/667895
