Find the volume of the solid obtained by rotating the region bounded by the given curves: y=x^3, y=0, x=1, about x=2. The answer is `(3pi)/5` but I can only get `(21pi)/10` .

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You need to use washers method to find the volume of the solid obtained by rotating the region bounded by the given curves, hence, you need to evaluate the definite integral such that:

`V = pi*int_0^1 (R^2(y) - r^2(y))dy`

R(y) represents the outer radius

`R(y) = 2 - root(3)(y) `

r(y) represents the inner radius

`r(x) = 2 - 1 = 1`

`V = pi*int_0^1 ((2 - root(3)y)^2 - 1^2)dy`

You need to expand the cube, such that:

`(2 - root(3)(y))^2 = 4 - 4root(3)y + root(3)(y^2) `

Using the property of linearity of integral, yields:

`V = pi*int_0^1 4dy - pi*int_0^1 4root(3)y dy + pi*int_0^1 root(3)(y^2) dy - pi*int_0^1 dy`

`V = pi(4y - 4(y^(1/3+1))/(1/3+1) + (3/5) y^(2/3+1) - y)|_0^1`

`V = pi(4y - 3y^(4/3) + (3/5)y^(5/3) - y)|_0^1`

Using the fundamental theorem of calculus yields:

`V = pi(4 - 3 + 3/5 - 1 - 0)|_0^1`

Reducing duplicate terms yields:

`V = pi*(3/5)`

**Hence, evaluating the volume of the solid obtained by rotating the region bounded by the given curves, yields` V = (3pi)/5` .**

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