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(16x^2)# and the x axis 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:
graph{(ysqrt(16x^2))=0 [8, 8, 2, 5]}
We need the point of intersection for the bounds of integration;
# sqrt(16x^2) = 0 => x =+ 4 # , and the range of#y# is#y in [4,4]#
By symmetry we can restrict
# y=sqrt(16x^2) => y^2=16x^2 => x=sqrt(16y^2)#
The Volume of Revolution about
# 1/2V = int_(y=a)^(y=b)2pi \ y \ g(y) \ dy #
# \ \ \ \ \ \ = int_(y=0)^(y=4)2pi \ y \ sqrt(16y^2) \ dy #
# \ \ \ \ \ \ = 2pi \ int_0^4 y \ sqrt(16y^2) \ dy #
We can use a substitution to perform this integral;
Let
#u=16y^2 => (du)/(dy)= 2y #
When# { (y=4), (y=4) :} => { (u=16), (u=0) :} #
Substituting into the integral we get:
# 1/2V = 2pi \ int_16^0 \ sqrt(u) (1/2)\ du #
# \ \ \ \ \ \ = pi \ int_0^16 \ sqrt(u) \ du #
# \ \ \ \ \ \ = pi \ [u^(3/2)/(3/2)]_0^16 #
# \ \ \ \ \ \ = 2/3pi \ [u^(3/2)]_0^16 #
# \ \ \ \ \ \ = 2/3pi \ (16^(3/2)  0) #
# \ \ \ \ \ \ = (128pi)/3 #
# :. V = (256pi)/3 #
Note:
The volume is that of a sphere of radius
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To 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{16  x^2} ) and the xaxis rotated about the xaxis, follow these steps:

Understand the Shell Method: The shell method involves integrating the circumference of cylindrical shells along the axis of rotation.

Determine the Bounds of Integration: Identify the interval over which the region is being revolved. In this case, it's from ( x = 4 ) to ( x = 4 ), as the function ( y = \sqrt{16  x^2} ) is defined for ( 4 \leq x \leq 4 ).

Set Up the Integral: The volume ( V ) of a solid generated by revolving a region bounded by ( y = f(x) ), the xaxis, and the lines ( x = a ) and ( x = b ) about the xaxis using the shell method is given by:
[ V = 2\pi \int_a^b x \cdot f(x) , dx ]
For this problem, we're revolving the region about the xaxis, so ( f(x) = \sqrt{16  x^2} ).

Evaluate the Integral: Plug in the bounds of integration and the function ( f(x) ), then integrate with respect to ( x ).
[ V = 2\pi \int_{4}^{4} x \cdot \sqrt{16  x^2} , dx ]

Simplify and Solve the Integral: This integral might require a trigonometric substitution or a usubstitution to solve. Once you find the antiderivative, evaluate it from ( x = 4 ) to ( x = 4 ) and subtract to find the final volume.

Check Units and Interpret the Result: Ensure that the units of volume are cubic units, as this represents a threedimensional quantity. Interpret the volume in the context of the problem.
By following these steps, you can use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the given region about the xaxis.
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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.
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