What is the Volume of Revolution if the area bounded by the curve #y=x^2-4x# and the #x#-axis is is rotated about the #x#-axis?

1 Answer
Steve M
Mar 5, 2017

# (512 pi) /15 \ unit^3#

Explanation:

I recommend that you always draw a sketch to clarify what needs calculating.

graph{x^2-4x [-10, 10, -5, 5]}

The curve intersects the #x#-axis, when:

# x^2-4x = 0 => x(x-4)=0 => x=0,4#

The Volume of Revolution about #Ox# is given by:

# VOR = int_alpha^beta \ piy^2 \ dx #

So in this case:

# VOR = int_0^4 \ pi \ (x^2-4x)^2 \ dx #
# " " = pi \ int_0^4 \ (x^4-8x^3+16x^2) \ dx #
# " " = pi [ x^5/5-2x^4+16x^3/3]_0^4 #
# " " = pi { (4^5/5-2*4^4+16*4^3/3)-(0) } #
# " " = pi (1024/5-512+1024/3)-(0) } #
# " " = (512 pi) /15 #

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