AP®MathsCollege BoardCalculus ABRevision NotesUnit 8: Applications of IntegrationVolumes with Cross SectionsSquares as Cross Sections

Squares as Cross Sections (College Board AP® Calculus AB)

Revision Note

Roger B

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Roger B

Expertise

Maths

Volumes with cross sections as squares

How can I find the volume of a solid with a square cross section?

  • Use the basic concept

    • If the area of the cross section of a solid is given by A open parentheses x close parentheses

      • and A open parentheses x close parentheses is continuous on open square brackets a comma space b close square brackets

    • Then the volume of the corresponding solid from x equals a to x equals b is

      • Volume equals integral subscript a superscript b A open parentheses x close parentheses space d x

  • You may need to create the cross sectional area function A open parentheses x close parentheses

    • E.g. to calculate the volume of a square-based right pyramid of height h and base side length 2 a

      • Consider a side view of the pyramid laid out along the x-axis:

Graph with x and y axes, a triangle, and the point (x, a - (a/h)x). Triangle vertices are at (0,a), (h,0), and (0,-a) with labeled lines and dashed vertical x line.

  • The line in red from a to h has the equation y equals a minus a over h x

  • At each value of x between 0 and h

    • the cross section of the pyramid is a square with side length 2 open parentheses a minus a over h x close parentheses

      • and area A open parentheses x close parentheses equals open parentheses 2 open parentheses a minus a over h x close parentheses close parentheses squared equals 4 a squared open parentheses 1 minus 1 over h x close parentheses squared

    • Therefore

table row Volume equals cell integral subscript a superscript h 4 a squared open parentheses 1 minus 1 over h x close parentheses squared space d x end cell row blank equals cell 4 a squared integral subscript a superscript h open parentheses 1 minus 2 over h x plus 1 over h squared x squared close parentheses space d x end cell row blank equals cell 4 a squared open square brackets x minus 1 over h x squared plus fraction numerator 1 over denominator 3 h squared end fraction x cubed close square brackets subscript 0 superscript h end cell row blank equals cell 4 a squared open parentheses open parentheses h minus 1 over h open parentheses h close parentheses squared plus fraction numerator 1 over denominator 3 h squared end fraction open parentheses h close parentheses cubed close parentheses minus 0 close parentheses end cell row blank equals cell 4 a squared times 1 third h end cell row blank equals cell 4 over 3 a squared h end cell end table

  • Alternatively, A open parentheses x close parentheses may depend on the values of another function given to you in the question

    • See the Worked Example

Worked Example

Let R be the region enclosed by the graph of f open parentheses x close parentheses equals 1 plus e to the power of negative x end exponent, the x- and y-axes, and the vertical line x equals 3, as shown in the figure below.

Graph showing the shaded region R under the curve y=1+e^(-x), bounded by the x-axis from 0 to 3 and the y-axis, with axes labeled x and y.

Region R is the base of a solid. For the solid, at each x the cross section perpendicular to the x-axis is a square. Find the volume of the solid.

Answer:

Use Volume equals integral subscript a superscript b A open parentheses x close parentheses space d x

At each x, the cross-sectional area is A open parentheses x close parentheses equals open square brackets f open parentheses x close parentheses close square brackets squared equals open parentheses 1 plus e to the power of negative x end exponent close parentheses squared

table row Volume equals cell integral subscript 0 superscript 3 open parentheses 1 plus e to the power of negative x end exponent close parentheses squared space d x end cell row blank equals cell integral subscript 0 superscript 3 open parentheses 1 plus 2 e to the power of negative x end exponent plus e to the power of negative 2 x end exponent close parentheses space d x end cell row blank equals cell open square brackets x minus 2 e to the power of negative x end exponent minus 1 half e to the power of negative 2 x end exponent close square brackets subscript 0 superscript 3 end cell row blank equals cell open parentheses 3 minus 2 e to the power of negative open parentheses 3 close parentheses end exponent minus 1 half e to the power of negative 2 open parentheses 3 close parentheses end exponent close parentheses minus open parentheses 0 minus 2 e to the power of negative open parentheses 0 close parentheses end exponent minus 1 half e to the power of negative 2 open parentheses 0 close parentheses end exponent close parentheses end cell row blank equals cell 3 minus 2 e to the power of negative 3 end exponent minus 1 half e to the power of negative 6 end exponent minus open parentheses negative 5 over 2 close parentheses end cell row blank equals cell 11 over 2 minus 2 e to the power of negative 3 end exponent minus 1 half e to the power of negative 6 end exponent end cell row blank equals cell 5.399186... end cell end table

The question doesn't specify units, so the units of volume will be units cubed

5.399 units3 (to 3 decimal places)

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    Roger B

    Author: Roger B

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.