Home > Calculus > Determining the Surface Area of a Solid of Revolution

What is the surface area of the solid created by revolving #f(x)=-2x^3-3x^2+6x-12# over #x in [2,3]# around the x-axis?

Answer 1

Use the surface area of rotation integral:
#SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx#
Integrate the following:
#SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx#

#SA = int_a^b 2pif(x) sqrt((1+((df(x))/dx)^2))dx# may look bad it is not really. Take the derivative of f(x) #(df(x))/dx = -6x^2 -6x+6 # now integrate

#SA =2pi int_2^3(-2x^3 - 3x^2 + 6x -12) sqrt(1+(-6x^2-6x+6))dx#

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Answer 2

Emma Aldridge

To find the surface area of the solid generated by revolving the function ( f(x) = -2x^3 - 3x^2 + 6x - 12 ) over the interval ( x ) in [2,3] around the x-axis, you can use the formula for surface area of revolution:

[ S = 2\pi \int_{a}^{b} |f(x)| \sqrt{1 + [f'(x)]^2} dx ]

Where: ( a = 2 ) and ( b = 3 ), ( f'(x) ) is the derivative of ( f(x) ).

First, find ( f'(x) ) and then plug it into the formula to evaluate the integral. Finally, multiply the result by ( 2\pi ) to get the surface area.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.