y = 5xe−x, y = 0, x = 3; about the y-axis

(a) Set up an integral for the volume V of the solid obtained by rotating the region bounded by the given curve about the specified axis.
(b) Use your calculator to evaluate the integral correct to five decimal places.

Relax

Respuesta :

Using the shell method, the volume is

[tex]V=\displaystyle2\pi\int_0^3x\cdot5xe^{-x}\,\mathrm dx=10\pi\int_0^3x^2e^{-x}\,\mathrm dx[/tex]

Integrate by parts, taking

[tex]u=x^2\implies\mathrm du=2x\,\mathrm dx[/tex]

[tex]\mathrm dv=e^{-x}\,\mathrm dx\implies v=-e^{-x}[/tex]

[tex]V=\displaystyle10\pi\left(-x^2e^{-x}\bigg|_0^3+2\int_0^3xe^{-x}\,\mathrm dx\right)[/tex]

[tex]V=\displaystyle-90\pi e^{-3}+20\pi\int_0^3xe^{-x}\,\mathrm dx[/tex]

Integrate by parts again, with

[tex]u=x\implies\mathrm du=\mathrm dx[/tex]

[tex]\mathrm dv=e^{-x}\,\mathrm dx\implies v=-e^{-x}[/tex]

[tex]V=-90\pi e^{-3}+20\pi\left(-xe^{-x}\bigg|_0^3\displaystyle\int_0^3e^{-x}\,\mathrm dx\right)[/tex]

[tex]V=-150\pi e^{-3}+20\pi\displaystyle\int_0^3e^{-x}\,\mathrm dx[/tex]

[tex]V=\displaystyle-150\pi e^{-3}+20\pi(-e^{-x})\bigg|_0^3[/tex]

[tex]\boxed{V=20\pi-170\pi e^{-3}}[/tex]

Answer:

a) [tex]2\pi\int_{0}^{3}x\cdot5xe^{-x}dx[/tex]

b) 36.24204 cubic units.

The shaded region will rotate about the y-axis.

x is from 0 to 3 from the graph.

Using the Shell method the volume is :

[tex]2\pi\int_{a}^{b}radius\cdot height\cdot dx\\=2\pi\int_{0}^{3}x\cdot5xe^{-x} dx\\=10\pi\int_{0}^{3}x^{2}e^{-x}dx\\=36.24204[/tex]

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