Question
find the volume of the solid generated by revolving around the x-axis: y=e^(x-8), y=0, x=0, x=10
So I know you can use the washer method, but the shell method can also be used and it should be better. The issue is I'm not certain how to actually do it. Any help is appreciated!
So I know you can use the washer method, but the shell method can also be used and it should be better. The issue is I'm not certain how to actually do it. Any help is appreciated!
Answers
washers:
v = ∫[0,10] πr^2 dx
where r = y = e^(x-8)
v = π∫[0,10] (e^(x-8))^2 dx
= π∫[0,10] e^(2x-16) dx
= π/2 e^(2x-16) [0,10]
= π/2 (e^4 - 1/e^16)
shells:
y = e^(x-8), so x = 8+lny
v = ∫[e^-8,e^2] 2πrh dy
where r = y and h = 10-x = 2-lny
v = 2π∫[e^-8,e^2] y(2-lny) dy
= π/2 (y^2(5-2lny)) [e^-8,e^2]
= π/2 (e^4 - 21/e^16)
Hmmm. better double-check my math. Anyway, that's the method.
v = ∫[0,10] πr^2 dx
where r = y = e^(x-8)
v = π∫[0,10] (e^(x-8))^2 dx
= π∫[0,10] e^(2x-16) dx
= π/2 e^(2x-16) [0,10]
= π/2 (e^4 - 1/e^16)
shells:
y = e^(x-8), so x = 8+lny
v = ∫[e^-8,e^2] 2πrh dy
where r = y and h = 10-x = 2-lny
v = 2π∫[e^-8,e^2] y(2-lny) dy
= π/2 (y^2(5-2lny)) [e^-8,e^2]
= π/2 (e^4 - 21/e^16)
Hmmm. better double-check my math. Anyway, that's the method.
Related Questions
the volume of the solid generated by revolving infinite region bounded by x-axis, x=k, and y=1/x+2 i...
Find the volume of the solid generated by revolving the region bounded by y= x^(1/2), y= 0, and x= 3...
Find the volume of the solid of revolution generated by revolving the region in the first quadrant b...
Find the volume of the solid of revolution generated by revolving the region in the first quadrant b...