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 $y = \sqrt{x}$ $y = sqrt(x$ ), $y = 0$ $y = 0$ , $y = 12 - x$ $y = 12 - x$ rotated about the x axis?

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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 $y = \sqrt{x}$ $y = sqrt(x$ ), $y = 0$ $y = 0$ , $y = 12 - x$ $y = 12 - x$ rotated about the x axis?

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Answer 1 · 2017-01-20T02:03:54

Volume = $\frac{99 π}{2}$ $(99pi)/2$

Explanation:

If you imagine an almost infinitesimally thin vertical line of thickness $δ x$ $deltax$ between the $x$ $x$ -axis and the curve at some particular $x$ $x$ -coordinate it would have an area:

$δ A \approx width \times height = y δ x = f (x) δ x$ $delta A ~~"width" xx "height" = ydeltax = f(x)deltax$

If we then rotated this infinitesimally thin vertical line about $O y$ $Oy$ then we would get an infinitesimally thin cylinder (imagine a cross section through a tin can), then its volume $δ V$ $delta V$ would be given by:

$δ V \approx 2 π \times radius \times thickness = 2 π x δ A = 2 π x f (x) δ x$ $delta V~~ 2pi xx "radius" xx "thickness" = 2pixdeltaA=2pixf(x)deltax$

If we add up all these infinitesimally thin cylinders then we would get the precise total volume $V$ $V$ given by:

$V=int_(x=a)^(x=b)2pi \ x \ f(x) \ dx$

Similarly if we rotate about $Ox$ instead of $Oy$ then we get the volume as:

$V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy$

So for this problem we have:

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We need the point of intersection for the bounds of integration;

$12-x = sqrt(x) => (12-x)^2 =x$
$:. 144-24x+x^2=x$
$:. x^2 - 25x + 144 =0$
$:. (x-9)(x-16) =0 => x=9,16$

And our solution is $x=9$ , (as $x=16$ is the solution for $-sqrt(x)$ )

$x=9 => y=12-x = 3$

So the bounding curves $y=sqrt(x)$ and $y=12-x$ meet at $(9,3)$

And we also need $x=g(y)$ , rather than $y=f(x)$

$y=12-x => x=12-y$
$y = sqrt(x) \ \ \ \ \ => x=y^2$

Then the required volume is given by:

$V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy$
$\ \ \= 2pi int_(y=0)^(y=3) \ y{(12-y) - (y^2)} \ dy$
$\ \ \= 2pi int_(y=0)^(y=3) \ y(12-y - y^2) \ dy$
$\ \ \= 2pi int_(y=0)^(y=3) \ 12y-y^2 - y^3) \ dy$
$\ \ \= 2pi [ 6y^2 - 1/3y^3 - 1/4y^4 ]_0^3$
$\ \ \= 2pi (54-9-81/4)$
$\ \ \= (99pi)/2$

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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 $y = \sqrt{x}$ $y = sqrt(x$ ), $y = 0$ $y = 0$ , $y = 12 - x$ $y = 12 - x$ rotated about the x axis?

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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 = sqrt(xy=√xy = sqrt(x), y = 0y=0y = 0, y = 12 - xy=12−xy = 12 - x rotated about the x axis?

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Explanation:

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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 $y = \sqrt{x}$ $y = sqrt(x$ ), $y = 0$ $y = 0$ , $y = 12 - x$ $y = 12 - x$ rotated about the x axis?