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Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis. y = 8x − x2 x = 0 y = 16

Respuesta :

LammettHash

Completing the square gives

[tex]y=8x-x^2=16-(x-4)^2[/tex]

and

[tex]16=16-(x-4)^2\implies(x-4)^2=0\implies x=4[/tex]

tells us the parabola intersect the line [tex]y=16[/tex] at one point, (4, 16).

Then the volume of the solid obtained by revolving shells about [tex]x=0[/tex] is

[tex]\displaystyle\pi\int_0^4x(16-(8x-x^2))\,\mathrm dx=\pi\int_0^4(x-4)^2\,\mathrm dx[/tex]

[tex]=\pi\dfrac{(x-4)^3}3\bigg|_{x=0}^{x=4}=\boxed{\dfrac{64\pi}3}[/tex]

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