1.
Să se afle volumul solidului
S
{\displaystyle S}
x
2
+
2
y
2
=
16
,
{\displaystyle x^{2}+2y^{2}=16,}
x
=
2
,
y
=
2
{\displaystyle x=2,\;y=2}
R.
Se observă că
S
{\displaystyle S}
z
=
16
−
x
2
−
2
y
2
{\displaystyle z=16-x^{2}-2y^{2}}
R
=
[
0
,
2
]
×
[
0
,
2
]
.
{\displaystyle R=[0,2]\times [0,2].}
Fubini
∬
R
(
16
−
x
2
−
2
y
2
)
d
A
=
∫
0
2
∫
0
2
(
16
−
x
2
−
2
y
2
)
d
x
d
y
=
{\displaystyle \iint _{\mathbf {R} }(16-x^{2}-2y^{2})dA=\int _{0}^{2}\int _{0}^{2}(16-x^{2}-2y^{2})dxdy=}
=
∫
0
2
(
16
x
−
x
3
3
−
2
y
2
x
)
|
0
2
d
y
=
∫
0
2
(
88
3
−
4
y
2
)
d
y
=
48.
{\displaystyle =\int _{0}^{2}(16x-{\frac {x^{3}}{3}}-2y^{2}x){\Bigg \vert }_{0}^{2}dy=\int _{0}^{2}\left({\frac {88}{3}}-4y^{2}\right)dy=48.}
2.
Să se calculeze integrala dublă:
∬
D
(
x
2
+
y
2
)
d
x
d
y
{\displaystyle \iint _{D}(x^{2}+y^{2})dxdy}
D
{\displaystyle D}
x
2
+
y
2
=
2
a
x
.
{\displaystyle x^{2}+y^{2}=2ax.}
R.
Ecuaţia domeniului
D
{\displaystyle D}
(
x
−
a
)
2
+
y
2
=
a
2
,
{\displaystyle (x-a)^{2}+y^{2}=a^{2},}
deci aceasta defineşte cercul cu centrul în punctul de coordonate
(
a
,
0
)
{\displaystyle (a,0)}
a
.
{\displaystyle a.}
Se face schimbarea de variabile
(
x
,
y
)
→
(
r
,
θ
)
{\displaystyle (x,y)\to (r,\theta )}
{
x
−
a
=
r
cos
θ
y
=
r
sin
θ
{\displaystyle {\begin{cases}x-a=r\cos \theta \\y=r\sin \theta \end{cases}}}
Noul domeniu de integrare (domeniul transformat ) este:
D
∗
=
{
(
x
,
y
)
∈
R
2
|
0
≤
r
≤
a
,
0
≤
θ
≤
2
π
}
.
{\displaystyle D^{*}=\{(x,y)\in \mathbb {R} ^{2}|0\leq r\leq a,\;0\leq \theta \leq 2\pi \}.}
Jacobianul acestei transformări este:
J
=
D
(
x
,
y
)
D
(
r
,
θ
)
=
|
∂
x
∂
r
∂
x
∂
θ
∂
y
∂
r
∂
y
∂
θ
|
=
|
cos
θ
−
r
⋅
sin
θ
sin
θ
r
⋅
cos
θ
|
=
r
cos
2
θ
+
r
sin
2
θ
=
r
,
{\displaystyle J={\frac {D(x,y)}{D(r,\theta )}}={\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}\end{vmatrix}}={\begin{vmatrix}\cos \theta &-r\cdot \sin \theta \\\sin \theta &r\cdot \cos \theta \end{vmatrix}}=r\cos ^{2}\theta +r\sin ^{2}\theta =r,}
iar
x
2
+
y
2
=
a
2
+
2
a
r
cos
θ
+
r
2
.
{\displaystyle x^{2}+y^{2}=a^{2}+2ar\cos \theta +r^{2}.}
Integrala de calculat devine:
∬
D
(
a
2
+
2
a
r
cos
θ
+
r
2
)
r
d
r
d
θ
=
∫
0
a
[
∫
0
2
π
(
a
2
r
+
2
a
r
2
cos
θ
+
r
3
)
d
θ
]
d
r
=
2
π
∫
0
a
(
a
2
r
+
r
3
)
d
r
=
3
π
a
4
2
.
{\displaystyle \iint _{D}(a^{2}+2ar\cos \theta +r^{2})rdrd\theta =\int _{0}^{a}\left[\int _{0}^{2\pi }(a^{2}r+2ar^{2}\cos \theta +r^{3})d\theta \right]dr=2\pi \int _{0}^{a}(a^{2}r+r^{3})dr={\frac {3\pi a^{4}}{2}}.}
![{\displaystyle R=[0,2]\times [0,2].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f249fefee2af0578a96baaa8b9f7173fd7b521fa)
![{\displaystyle \iint _{D}(a^{2}+2ar\cos \theta +r^{2})rdrd\theta =\int _{0}^{a}\left[\int _{0}^{2\pi }(a^{2}r+2ar^{2}\cos \theta +r^{3})d\theta \right]dr=2\pi \int _{0}^{a}(a^{2}r+r^{3})dr={\frac {3\pi a^{4}}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/787934acdbc9fd1dd229afd6cba9840792258958)