How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region $y = x^3$ , $y = 0$ , $x = 2$ rotated about the x axis?

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Answer 1 · 2015-05-23T10:18:25

The answer is $V = (128pi)/7$ .

Firstly, you should graph your three functions to see clearly what is the delimited region :

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We are rotating our plane region around the x-axis, which means we have to write our function $f(x) = x^3$ in the form $f(y) =root(3)(y)$ .

The radius of any cylinder in our final volume will be given by $y$ .

And the width will be $2-root(3)(y)$ .

(Look here for more details or diagrams about the cylinders)

Therefore, the cross-sectional area of any cylinder will be :

$A(y) = 2piy(2-root(3)(y)) = 4piy - 2piy^(4/3)$

The radius of the cylinders, according to the delimited region, goes from :

$f(x) = 0 = y$ (with $x = 0$ ) to $f(x) = 8 = y$ (with $x=2$ ).

So the volume of our solid, which is the sum of all the cross-sectional area of the cylinders, is :

$V = int_0^8A(y)dy$ .

The antiderivative of $ay^n$ is given by $a*1/(n+1)y^(n+1)$ .

Thus, the antiderivative of $A(y)$ is :

$1/2 4piy^2 - 3/7 2piy^(7/3) = 2piy^2(1-3/7root(3)y)$

Now we can calculate the integral :

$V = int_0^8A(y)dy = [2piy^2(1-3/7root(3)y)]_0^8$

$= 2pi [y^2(1-3/7root(3)y)]_0^8 = 2pi(8^2(1-(3*2)/7)-0)$

$= 2pi*64/7 = (128pi)/7$ .

How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y = x^3 y = x^3, y = 0y = 0, x = 2x = 2 rotated about the x axis?