Resposta:
Temos o volume da esfera em fun\cc
a
˜
o de seu raio:
V(r)=
3
4
πr
3
Deseja-se calcular a taxa de varia\cc
a
˜
o m
e
ˊ
dia de V, para
1≤r≤3.
\begin{gathered}\large\begin{array}{l} \textsf{A taxa de varia\c{c}\~ao m\'edia \'e dada por}\\\\ \mathsf{\overline{V}\big|_{r_1\to r_2}=\dfrac{V(r_2)-V(r_1)}{r_2-r_1}}\\\\ \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{V(3)-V(1)}{3-1}}\\\\ \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{\frac{4}{3}\pi \cdot 3^3-\frac{4}{3}\pi \cdot 1^3}{3-1}} \end{array}\end{gathered}
A taxa de varia\cc
a
˜
o m
e
ˊ
dia
e
ˊ
dada por
V
∣
∣
∣
r
1
→r
2
=
r
2
−r
1
V(r
2
)−V(r
1
)
V
∣
∣
∣
1→3
=
3−1
V(3)−V(1)
V
∣
∣
∣
1→3
=
3−1
3
4
π⋅3
3
−
3
4
π⋅1
3
\begin{gathered}\large\begin{array}{l} \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{\frac{4}{3}\pi \cdot (3^3-1^3)}{3-1}}\\\\ \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{4\pi}{3}\cdot \dfrac{27-1}{3-1}}\\\\ \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{4\pi}{3}\cdot \dfrac{26}{2}}\\\\ \mathsf{\overline{V}\big|_{1\to 3}=\dfrac{4\pi}{3}\cdot 13}\\\\ \boxed{\begin{array}{c}\mathsf{\overline{V}\big|_{1\to 3}=\dfrac{52\pi}{3}\quad\dfrac{u.v.}{u.c.}} \end{array}}\quad\longleftarrow\quad\textsf{esta \'e a resposta.} \end{array}\end{gathered}
V
∣
∣
∣
1→3
=
3−1
3
4
π⋅(3
3
−1
3
)
V
∣
∣
∣
1→3
=
3
4π
⋅
3−1
27−1
V
∣
∣
∣
1→3
=
3
4π
⋅
2
26
V
∣
∣
∣
1→3
=
3
4π
⋅13
V
∣
∣
∣
1→3
=
3
52π
u.c.
u.v.
⟵esta
e
ˊ
a resposta.
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