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 = 0#, #y = 12  x# rotated about the x axis?
Volume =
If you imagine an almost infinitesimally thin vertical line of thickness
#delta A ~~"width" xx "height" = ydeltax = f(x)deltax#
If we then rotated this infinitesimally thin vertical line about
#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=int_(x=a)^(x=b)2pi \ x \ f(x) \ dx #
Similarly if we rotate about
# V=int_(y=a)^(y=b)2pi \ y \ g(y) \ dy #
So for this problem we have:
We need the point of intersection for the bounds of integration;
# 12x = sqrt(x) => (12x)^2 =x #
# :. 14424x+x^2=x #
# :. x^2  25x + 144 =0 #
# :. (x9)(x16) =0 => x=9,16 #
And our solution is
# x=9 => y=12x = 3 #
So the bounding curves
And we also need
# y=12x => x=12y #
# 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{(12y)  (y^2)} \ dy #
# \ \ \= 2pi int_(y=0)^(y=3) \ y(12y  y^2) \ dy #
# \ \ \= 2pi int_(y=0)^(y=3) \ 12yy^2  y^3) \ dy #
# \ \ \= 2pi [ 6y^2  1/3y^3  1/4y^4 ]_0^3 #
# \ \ \= 2pi (54981/4) #
# \ \ \= (99pi)/2 #
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To use the shell method to set up and evaluate the integral for the volume of the solid generated by revolving the given plane region about the xaxis, we follow these steps:
 Determine the limits of integration by finding the intersection points of the curves involved. In this case, we find the intersection point of (y = \sqrt{x}) and (y = 12  x).
(12  x = \sqrt{x})
Solve for (x) to find the intersection point.

Decide whether to integrate with respect to (x) or (y). Since the problem specifies revolving about the xaxis, we'll integrate with respect to (x).

Determine the radius of the shells. In this case, the radius is the distance from the axis of rotation (xaxis) to the outer curve (12  x).

Determine the height of the shells. This is the difference between the upper and lower curves at a given (x)value.

Set up the integral for the volume using the shell method formula:
[ V = 2\pi \int_{a}^{b} r(x)h(x) , dx ]
Where (a) and (b) are the limits of integration, (r(x)) is the radius of the shell, and (h(x)) is the height of the shell.

Evaluate the integral using the determined limits of integration and the functions for (r(x)) and (h(x)).

Calculate the result.

Include appropriate units if necessary.
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When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
When evaluating a onesided 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.
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