How do you find the volume of the solid #y=x^2, y=x^3# revolved about the x-axis?

1 Answer
Steve M
Jan 18, 2017

Volume # (2pi)/35 " unit^3#

Explanation:

graph{(y-x^2)(y-x^3)=0 [-0.1, 1.5, -0.1, 1.5]}

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

# V= int_(x=a)^(x=b) \ pi y^2 \ dx #

So, the volume of revolution bounded by two curves #f(x)# and #g(x)# with #f(x) gt g(x)# is given by the difference between the individual solids of revolution, thus:

# V = int_(x=a)^(x=b) \ pi f^2(x) \ dx - int_(x=a)^(x=b) \ pi g^2(x) \ dx #
# \ \ = int_(x=a)^(x=b) \ pi {f^2(x) - g^2(x)} \ dx #

We can see by observation that the points of intersection of the curves #y=x^2# and #y=x^3# are #(0,0)# and #(1,1)#.

So for for this problem:

# V= int_0^1 \ pi { (x^2)^2 -(x^3)^2} \ dx #
# \ \ \= pi \ int_0^1 \ x^4-x^6 \ dx #
# \ \ \= pi \ [x^5/5-x^7/7]_0^1#
# \ \ \= pi \ { (1/5-1/7) - (0)}#
# \ \ \= pi * 2/35#

# \ \ \= (2pi)/35#

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