What is the volume of the solid produced by revolving #f(x)=x^2+3x-sqrtx, x in [0,3] #around the x-axis?

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Answer 1 · 2016-12-30T09:31:30

601.5 cubic units, nearly

The two x-scaled and y-scaled graphs reveal that the said

shuttlecock-like solid of revolution has two parts, having just the x-

intercept point #(0.1_ -, 0)# as a common point, The very small part

near the origin might appear as a knot. This one has a convex

surface, in contrast to the other that has concave surface.

The volume is

#pi int (x^2+3x-sqrtx)^2 dx#, for x from 0 to 3.

#=pi int ( (x^4 +9x^2+x-2(3x)(sqrtx)-2(sqrtx)(x^2)+2(x^2)(3x)) dx,#

for the limits

#=pi[x^5/5+9(x^3/3)+x^2/2-6(2/5x^(5/2))-2(2/7x^(7/2)+6(x^4/4)],#

between x =. 0 and 3

#=pi[243/5+81 +9/2-sqrt 3 (108)(1/5+1/7)+243/2]#

#=601.5 cubic units, nearly.

graph{0.1(x^2+3x-sqrtx) [-10, 10, -5, 5]}
graph{(10(x^2+3x-sqrtx)) [-0.1, 0.1, -10, 10]}