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Find the volume of the solid obtained by rotating about y axis the region between y=x...

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yapayapa | Honors

Posted September 13, 2013 at 3:38 PM via web

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Find the volume of the solid obtained by rotating about y axis the region between y=x and y=x^2.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted September 13, 2013 at 3:57 PM (Answer #1)

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You may use the shell method to evaluate the volume of the requested volume, such that:

`V = int_a^b 2pi*x*f(x) dx`

`2pi*x` represents the circumference of the shell

`f(x)` represents the height of the shell

You need to find the limits of integration a,b, hence, you need to solve the following system of equations, such that:

`{(y = x),(y = x^2):} => x = x^2 => x^2 - x = 0 => x(x - 1) = 0`

Using zero product rule, yields:

`x = 0`

`x - 1 =0 => x = 1`

Hence, evaluating the limits of integration yields `a = 0, b = 1` .

Identifying the height `f(x) = x - x^2` , you may evaluate the volume, such that:

`V = int_0^1 (2pi*x)*(x - x^2) dx `

Taking out the constant 2pi, yields:

`V = 2pi*int_0^1 (x^2 - x^3)dx`

Using the property of linearity of integral, yields:

`V = 2pi*(int_0^1 x^2 - int_0^1 x^3) dx`

`V = 2pi*(x^3/3 - x^4/4)|_0^1`

By fundamental theorem of calculus, yields:

`V = 2pi*(1^3/3 - 1^4/4 - 0^3/3 + 0^4/4)`

`V = 2pi*(1/3 - 1/4) => V = 2pi*1/12 => V = pi/6`

Hence, evaluating the volume of the solid obtained by rotating about y axis the region between `y=x` and `y=x^2` , yields `V = pi/6.`

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