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=2x^2+5#, #y=x+3#, the y-axis, and the line #x=3# rotated about the x-axis?

Answer 1

See the explanation section, below.

Here is a graph of part of the region to be rotated about the #x# axis.

In order to use shells, we must take our representative slice parallel to the axis of rotation. In this case, that's the #x# axis, so the thickness of our shells will be #dy#.

That means we'll need to rewrite the curves as functions of #y#

#x=sqrt((y-5)/2)#, #x=y-3#

Every shell will have #"radius" =y#, #"thickness"=dy# and #"height" = x_"right" - x_"left"#

As the graph shows, there are three separate integrals we need to do, because the calculation of #x_"right"# and #x_"left"# change as #y# goes from #3# (the lower bound) to #23# (the uppr bound, not on my graph).

(Are we sure we want to use shells for this?)

From #y=3# to #y=5# , the height of the cylindrical shell will be
#"height" = x_"right" - x_"left" = (y-3)-0 = y-3#

So we need to integrate

#int_3^5 2piy(y-3)dy = 2pi int_3^5 (y^2-3y)dy# which is left to the student.

From #y=5# to #y=6#, the height of the cylindrical shell will be
#"height" = x_"right" - x_"left" = (y-3)- sqrt((y-5)/2)#

So we need to integrate

#int_5^6 2piy(y-3- sqrt((y-5)/2))dy = 2pi int_5^6 (y^2-3y- ysqrt((y-5)/2))dy# the last term of which can be integrated by substitution (or parts).

From #y=6# to #y=23#, the height of the cylindrical shell will be
#"height" = x_"right" - x_"left" = 3- sqrt((y-5)/2)#

So we need to integrate

#int_6^23 2piy(3- sqrt((y-5)/2))dy = 2pi int_6^23 (3y- ysqrt((y-5)/2))dy# the last term of which can be integrated by substitution (or parts).

Washers

To use washers take the slices perpendicular to the axis of rotation.

As x varies from #0# to #3#, the greater radius at #x# is #R = 2x^2+5# and the lesser radius is #r = x+3#,

We need to integrate

#int_0^3 pi (R^2-r^2)dx = pi int_0^3 ((2x^2+5)^2-(x+3)^2)dx# which is simply the integral of a degree 4 polynomial when we expand it.

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Answer 2

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, you first need to determine the limits of integration and the radius and height of the shells.

  1. Find the points of intersection of the curves (y = 2x^2 + 5) and (y = x + 3): (2x^2 + 5 = x + 3) (2x^2 - x + 2 = 0) Solve for (x) to find the points of intersection.

  2. The limits of integration will be from the x-coordinate of the leftmost point of intersection to the x-coordinate of the rightmost point of intersection.

  3. For each shell, the radius is the x-coordinate of the shell, and the height is the difference between the y-coordinates of the two curves at that x-coordinate.

  4. The volume of each shell is (2\pi rh), where (r) is the radius and (h) is the height. Integrate this expression over the limits of integration to find the total volume.

The integral to evaluate will look like:

[\int_{a}^{b} 2\pi x (y_{outer} - y_{inner}) , dx]

where (a) and (b) are the x-coordinates of the points of intersection, and (y_{outer}) and (y_{inner}) are the equations of the outer and inner curves, respectively.

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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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