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Sagot :
To calculate the total volume of the solid we will divide it into n shell cyline thickness of [tex]$\Delta y$.[/tex]
The internal shell cylinder circumference C shall be[tex]$2 \pi r_i$[/tex]
where r is the in The shell cylinder internal area A shall be C.
where h is the height
The volume of the shell cylinder[tex]$\Delta v$[/tex] shall be[tex]$A \cdot \Delta y$.[/tex]
Then we will have:
[tex]\Delta v=A \cdot \Delta y=C \cdot h \cdot \Delta y=2 \pi r_i h \cdot \Delta y \Rightarrow d v=2 \pi r_i h \cdot d y$[/tex]
The total volume of the solid shall be the summation of all n shell cylinders
[tex]\int d v=\int 2 \pi r_i h d y \Rightarrow V=\int_a^b 2 \pi r_i h d y \Rightarrow V=2 \pi \int_a^b r_i h d y$[/tex]
[tex]\int d v=\int 2 \pi r_i h d y \Rightarrow V=\int_a^b 2 \pi r_i h d y \Rightarrow V=2 \pi \int_a^b r_i h d y[/tex]
[tex]The value $a$ is the value of $y$ at the point where $x=0$ then:For $x=0 \Rightarrow y=0^3 \Rightarrow y=0$ then $a=0$.The value $b$ is the value of $y$ at the point where $x=1$ then:[/tex]
[tex]For $x=1 \Rightarrow y=1^3 \Rightarrow y=1$ then $b=1$.Since the integration will be with variable $y$ we have to express the function $y=x^3$ in terms of $y$ :$[/tex]
y=x^3 \[tex]Rightarrow x=\sqrt[3]{y}[/tex]
Substituting in V, we will have:
[tex]V= & 2 \pi \int_a^b r_i h d y=2 \pi \int_0^1(1-y)(1-\sqrt[3]{y}) d y=2 \pi[/tex]\[tex]int_0^1(1-y)\left(1-y^{\frac{1}{3}}\right) d y \\[/tex]
&[tex]1-y \\& \frac{1-y^{\frac{1}{3}}}{1-y} \\& \frac{-y^{\frac{1}{3}}+y^{\frac{1}{3}} \cdot y^{\frac{1}{1}}}{} \\[/tex]
[tex]& 1-y-y^{\frac{1}{3}}+y^{\frac{1}{3}+\frac{1}{1}}= \\& =1-y-y^{\frac{1}{3}}+y^{\frac{1+3}{3}}= \\[/tex]
[tex]& =1-y-y^{\frac{1}{3}}+y^{\frac{4}{3}}= \\& =y^{\frac{4}{3}}-y^{\frac{1}{3}}-y+1[/tex]
Substituting in V:
[tex]V=2 \pi \int_0^1\left(y^{\frac{4}{3}}-y^{\frac{1}{3}}-y+1\right) d y=2[/tex][tex]{1}{3}+1}}{\frac{1}{3}+1}-\frac{y^{1+1}}[/tex]{[tex]1+1}+y\right]_0^1=2[/tex] [tex]\pi\left[\frac{y^{\frac{4+3}{3}}}{\frac{4+3}{3}}[/tex]-[tex]\frac{y^{\frac{1+3}{3}}}{\frac{1+3}{3}}-\frac{y^2}{2}+y\right]_0^1$[/tex]
Calculating the definite integral:
By definition[tex], $\int_a^b f(x) d x=\left.F(x)\right|_a ^b=F(b)-F(a)$[/tex] then:
[tex]& V=\frac{\pi}{14}\left[\left(12 \cdot 1^{\frac{7}{3}}-21 \cdot 1^{\frac{4}{3}}-14[/tex][tex]\cdot 1^2+28 \cdot 1\right)-\left(12 \cdot 0^{\frac{7}{3}}-21 \cdot[/tex] [tex]0^{\frac{4}{3}}-14 \cdot 0^2+28 \cdot 0\right)\right] \\[/tex]
& V=[tex]\frac{\pi}{14}[(12 \cdot 1-21 \cdot 1-14 \cdot 1+28 \cdot 1)-\underbrace{(12 \cdot 0-21 \cdot 0-14 \cdot 0+28 \cdot 0)}_0][/tex]
[tex]& V=\frac{\pi}{14}(12-21-14+28)=\frac{\pi}{14}(12+28-21-14)=\frac{\pi}{14}(40-35)=\frac{5 \pi}{14} \approx \frac{5 \cdot 3,1416}{14} \approx \frac{15,708}{14} \\& V \approx 1,122[/tex]
Answer:[tex]$V \approx 1,122 u . v$. (units of volume)[/tex]
To learn more about total volume visit:
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