How do you find the volume of the solid obtained by rotating the region bounded by the curves #f(x) = 3x^2# and #f(x) = 5x+2 # about the x axis?

1 Answer
bp
Aug 20, 2015

#317617/810 pi#

Explanation:

The region bounded by the two functions, a vertical parabola and a straight line is shown in the picture. On solving the two equations the points of intersection can be easily found to be# (-1/3, 1/3)# and (2,12)

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If #y_1= 3x^2# and #y_2#= 5x+2, consider an element of length #y_2 -y_1# and width dx, of the region bounded by the two functions. If this element is rotated about x axis, the volume of the elementary disc so formed would be #pi(y_2-y_1)^2# dx.

The volume of the solid formed by rotation of the whole region, about x axis would be

#int_(-1/3)^ 2 pi(y_2-y_1)^2 dx#

#int_(-1/3) ^2 pi(5x+2-3x^2)^2 dx#

#int_(-1/3)^ 2 pi(9x^4 -30x^3 + 13x^2 +20x +4)dx#

On solving, this integral would work out to be=#317617/810 pi#

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