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

Dec 30, 2016

601.5 cubic units, nearly

Explanation:

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 $\left({0.1}_{-} , 0\right)$ 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

$\pi \int {\left({x}^{2} + 3 x - \sqrt{x}\right)}^{2} \mathrm{dx}$, 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

$= \pi \left[\frac{243}{5} + 81 + \frac{9}{2} - \sqrt{3} \left(108\right) \left(\frac{1}{5} + \frac{1}{7}\right) + \frac{243}{2}\right]$

#=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]}