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=6x^2#, #y=6sqrtx# rotated about the y-axis?

Answer 1

The normal revolution method calls for:

#V = pir^2h = sum_(a=a_0)^b pi(r(x))^2Deltax = piint_a^b (r(x))^2dx#
where one stacks circle analogs of varying radius #r(x)# across an orthogonal interval #Deltax#.

In contrast, the shell method calls for a volume formula as such:

#V = int_a^b 2pixr(x)dx#
where the way #r(x)# varies determines the height and shape of the revolved solid. Here, #Deltax# is instead the thickness of the shell and #x# is its radius. This is sometimes easier, especially if you are rotating about the y-axis instead of the x-axis.

Let's see how this looks.

#y = 6x^2#: graph{6x^2 [-2, 2, -1, 1]}
#y = 6sqrtx#: graph{6sqrtx [-2, 2, -2, 6]}
If you layer these graphs on top of each other, you can see that they intersect to form a "stretched lemon" of sorts. Let's find where they intersect to determine our #Deltax#.
Besides #x = 0#:
#6x^2 = 6sqrtx# #x^2 = sqrtx# #x^4 = x# #x^3 = 1# #x = 1#
Thus, the practical interval is #[0, 1]#, as you can see here.
So, taking the area between the two curves as the difference between the top #(6sqrtx)# and bottom #(6x^2)# curves and using it as #r(x)#:
#int_0^1 2pixr(x)dx#
#= 2piint_0^1 x[6sqrtx - 6x^2]dx#
#= 12piint_0^1 x[sqrtx - x^2]dx#
#= 12piint_0^1 x^(3/2) - x^3dx#
#= 12pi [2/5x^(5/2) - x^4/4]|_(0)^(1)#
#= 12pi [(2/5(1)^(5/2) - (1)^4/4) - (2/5(0)^(5/2) - (0)^4/4)]#
#= 12pi (2/5 - 1/4)#
#= 12pi (8/20 - 5/20)#
#= 12pi (3/20)#
#= 36/20 pi#
#= color(blue)(9/5 pi ~~ 5.6549 "u"^3)#
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Answer 2

To use the shell method for finding the volume of the solid generated by revolving the region ( y = 6x^2 ) and ( y = 6\sqrt{x} ) about the ( y )-axis, follow these steps:

  1. Determine the limits of integration by finding the intersection points of the curves ( y = 6x^2 ) and ( y = 6\sqrt{x} ).

  2. Express the differential volume element ( dV ) in terms of ( y ) and ( dy ).

  3. The integral to find the volume using the shell method is: [ V = 2\pi \int_{a}^{b} y \cdot \text{radius} \cdot dy ] where ( a ) and ( b ) are the lower and upper limits of integration respectively, ( y ) is the variable of integration, and the radius is the distance from the axis of rotation to the outer edge of the solid.

  4. The radius can be found by expressing ( x ) in terms of ( y ) and finding the distance from the ( y )-axis to the curve.

  5. Evaluate the integral using the determined limits of integration and the appropriate expression for the radius.

Following these steps, the integral should give the volume of the solid generated by revolving the given region about the ( y )-axis using the shell method.

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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