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?

Question

Cameron Alexander · Accepted Answer

Volume = #(99pi)/2 #

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

#delta A ~~"width" xx "height" = ydeltax = f(x)deltax#

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

#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# 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:

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 #

Charles Arocha · Answer

undefined 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 x-axis, we follow these steps:

1. 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.

2. Decide whether to integrate with respect to $x$ or $y$. Since the problem specifies revolving about the x-axis, we'll integrate with respect to $x$.

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

4. Determine the height of the shells. This is the difference between the upper and lower curves at a given $x$-value.

5. 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.

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

7. Calculate the result.

8. Include appropriate units if necessary.