Volumes of Solids

Example 1: Cross Sectional Areas

The solid lies between planes perpendicular to the x-axis at x = 2 to x = 2 . The cross-sections perpendicular to the x-axis are semi-circles whose diameter run from g ( x ) = x 2 to f ( x ) = x 2 + 8 . Find the volume of the solid.

Step 1: Identify the shape whose area equation, A(x) , you need find, and write down that general area formula.

 

In this problem you see that the problem talks about semi-circles , and so you know this is the general area formula you need to write down.

“The cross sections perpendicular to the x-axis are semi-circles

          General Area of a Circle Equation: A = π r 2

          Derive, General Area of a Half-Circle ( semi-circle ): A = 1 2 π r 2

Step 2: Use your given equations to create an equation for the edge (base, height) or dimension (diagonal, diameter) of the geometric shape you are working with.

 

First, you will want to sketch a graph if one is not given.

Once you sketch out your two given equations g ( x ) = x 2 to f ( x ) = x 2 + 8 , draw a line between the two graphs to represent the diameter that was described in the language of the problem.

 

Second , you determine how to find an equation for the diameter you just drew.

When you are looking to find the distance between two curves you follow the same process that you learned when you were finding the Area Between Two Curves.

 

The Distance Between 2 Curves = Top Bottom = Diameter

 

What you have found is a formula for the diameter of the semi-circle .

“…whose diameter run from g ( x ) = x 2 to f ( x ) = x 2 + 8 .

 

 

The Edge: Diameter

 

Distance Between 2 Curves =

Top Bottom = Diameter

Top = x 2 + 8

Bottom = x 2

Diameter = Top Bottom

Diameter = x 2 + 8 ( x 2 )

Diameter = x 2 + 8 x 2

Diameter = 2 x 2 + 8

The Shape: Semi-circle

Step 3: Take your general area formula from Step 1 , and determine equations for each variable of the formula based on the edge equation you found in Step 2 .

 

Here you know you only need to plug a radius into your general area formula, but you currently have an equation for the diameter .

 

At this point you remind yourself that the radius of a circle equals half of the diameter .

radius = 1 2 diameter

Try not to think too hard about this. You have an equation for the diameter from Step 2 . Plug that diameter equation in for the diameter of your radius equation, and you now have an equation for the radius .

 

You can then simplify this equation by distributing the 1 2 before using it in Step 4 .

General Area Formula from Step 1 :  Area of a Half-Circle ( semi-circle ) = A = 1 2 π r 2

Edge equation from Step 2 : Diameter = 2 x 2 + 8

 

radius = 1 2 diameter

radius = 1 2 ( 2 x 2 + 8 )

radius = x 2 + 4

Step 4: Create your actual area equation, A(x) , using the pieces you found in Step 3 .

 

Here you take the radius equation you created in Step 3 and plug it in for the radius in general area formula from Step 1 .

 

You now have the actual area equation, A(x) , that you need for your Cross-Sectional Area formula.

 

You have the formula for one slice of the loaf of bread.

General Area Formula from Step 1 :  Area of a Half-Circle ( semi-circle ) = A = 1 2 π r 2

Radius equation from Step 3 : radius = x 2 + 4

 

A ( x ) = 1 2 π ( x 2 + 4 ) 2 = One Slice

 

Step 5: Determine the bounds of the integral, a b .

 

Here the language of the problem gives you the bounds you need for your Cross-Sectional Area formula.

“The solid lies between planes perpendicular to the x-axis at x = 2 to x = 2 .”

 

a b = 2 2

 

Step 6: Setup and evaluate the definite integral you created, Volume = a b A ( x ) dx , to find your final Volume.

 

Area equation, A(x) , from Step 4 : A ( x ) = 1 2 π ( x 2 + 4 ) 2

Bounds from Step 5 : 2 2

Volume = 2 2 1 2 π ( x 2 + 4 ) 2 dx

Expand the squared quantity:

2 2 1 2 π ( x 2 + 4 ) ( x 2 + 4 ) dx =

Multiply the quantities using distribution:

2 2 1 2 π ( x 4 4 x 2 4 x 2 + 16 ) dx =

2 2 1 2 π ( x 4 8 x 2 + 16 ) dx =

Distribute 1 2 π to the entire quantity:

2 2 1 2 π x 4 4 π x 2 + 8 π dx =

Apply the Definite Integral Power Rule:

1 2 π x 5 5 4 π x 3 3 + 8 π x | 2 2 =

 

[ 1 2 π ( 2 ) 5 5 4 π ( 2 ) 3 3 + 8 π ( 2 ) ] [ 1 2 π ( 2 ) 5 5 4 π ( 2 ) 3 3 + 8 π ( 2 ) ] = 256 15 π

 

2 2 1 2 π ( x 2 + 4 ) 2 dx = 256 15 π = Volume of Solid = Volume of Full Loaf of Bread

Final Result:

The volume of the solid (loaf of bread) created by semi-circle cross sectional areas (slices of bread) that are perpendicular to the x-axis and whose diameter run from g ( x ) = x 2 to f ( x ) = x 2 + 8 is 256 15 π .

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