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(16-x^2)# and the x axis rotated about the x axis?

Answer 1

Volume #= (256pi)/3 \ "unit"^3#

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:
graph{(y-sqrt(16-x^2))=0 [-8, 8, -2, 5]}

We need the point of intersection for the bounds of integration;

# sqrt(16-x^2) = 0 => x =+- 4 #, and the range of #y# is #y in [-4,4]#

By symmetry we can restrict #y in [0,4]# and double the result.

# y=sqrt(16-x^2) => y^2=16-x^2 => x=sqrt(16-y^2)#

The Volume of Revolution about #Ox# is given by:

# 1/2V = int_(y=a)^(y=b)2pi \ y \ g(y) \ dy #
# \ \ \ \ \ \ = int_(y=0)^(y=4)2pi \ y \ sqrt(16-y^2) \ dy #
# \ \ \ \ \ \ = 2pi \ int_0^4 y \ sqrt(16-y^2) \ dy #

We can use a substitution to perform this integral;

Let #u=16-y^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 #4#, so we can validate the volume as #V=4/3pir^3 = (256pi)/3#

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Answer 2

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 x-axis rotated about the x-axis, follow these steps:

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

  2. 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 ).

  3. Set Up the Integral: The volume ( V ) of a solid generated by revolving a region bounded by ( y = f(x) ), the x-axis, and the lines ( x = a ) and ( x = b ) about the x-axis 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 x-axis, so ( f(x) = \sqrt{16 - x^2} ).

  4. 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 ]

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

  6. Check Units and Interpret the Result: Ensure that the units of volume are cubic units, as this represents a three-dimensional 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 x-axis.

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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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