Answer:
Step-by-step explanation:
[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]
y = 0, x = 0 and x = 3
Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by
[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]
[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]
Integrate the above integral over the limits x=0 to x=3 which implies
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]
Solve by subsititution
[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]
Also, apply the new limits
[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]
This implies,
[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]
[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Let,
[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]
Integrate by parts the above integral
[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]
Again integrate by parts
[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]
Integrate by parts formula
[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]
Therefore,
[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]
This implies, the volume is
[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]
That is, up to three decimal places
[tex]V\approx 2.118[/tex]
