Slice It, Formulate Area, Integrate Over the 3rd Dimension

Here's what we're doing. We're going to find the volume of a 3-dimensional solid by cutting it into an infinite number of similar slices or wafers. Now, volume is area of the base times the height, so we're going to find a function for the area of the base of a typical slice and then we'll sum (integrate) it over the 3rd dimension to find volume. The dx of the integral will serve as the thickness or height of the slice or wafer.

We'll start with a bread since it's something we slice all the time.

Example 1: Find the volume of a bread (shown) with base represented by the ellipse
with equation x ² + 16y² = 16, if each slice cut perpendicular to the base is a semi-circle.

Solution: x² + 16y² = 16 is an ellipse with a = 4 and b = 1, so, as shown, the slices will extend
from – 4 to + 4 along the x-axis. This means we'll integrate with respect to x and therefore, we need the formula for the area of a "typical" slice in terms of x.

Since each slice is a semi-circle with radius = y and the area of a semi-circle is , we can substitute y for r to get . However, the slices pile up along the x-axis from – 4 to + 4 so we'll do some algebra to find a substitution expression for y² in terms of x.

Now, because the elliptical base is symmetric to the origin, rather than integrating the area formula from – 4 to + 4, we'll take double the integral from zero to + 4 to find the volume.

When we substitute for y² in the area formula, we get:

Now, we'll take double the integral from zero to + 4. So, the volume of the bread is

Now we bring out and combine the constants, then we integrate to get

The volume of the bread is cubic units.

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Sometimes, we need to use similar figures to formulate the Area function in terms of the correct variable of integration. Here's the classic pyramid example you'll find in every Calculus text book in the universe.

Example 2: Find the volume of a right pyramid if the altitude = h and the length of a side of the square base = a. (Remember that h and a are real number measures -- they're constants.)

Solution: We place the pyramid with its vertex at the origin so that the x-axis forms the axis of symmetry of the pyramid and it extends out from the center point of the base (as shown).

The purple square represents a "typical" slice. It is x-units right of the vertex and since it extends from y-units above the axis to y-units below it, each side is 2y units long.

Remember that a and h are constants. The area of the purple square located x units from the origin, is 4y², since it is 2y units per side. We need to integrate 4y² from 0 to h but the slices pile up along the x-axis, so we need it in terms of x, a, and h -- not in terms of y. We compare the dimensions of the purple square with those of the base square to get:

Now we can write A( y) = 4y² in terms of x using this substitution for y. We get

Now we integrate this from x = 0 to x = h and we'll have the volume of the pyramid. Recall that a and h are constants -- real number values for the dimensions of the pyramid.

Had we been told that the base of the pyramid is a square 9 meters per side and altitude is 10 meters, we would substitute a = 9 and h = 10 to get the volume. We now have a formula for the volume of a right square pyramid with base side = a and altitude = h. To find the volume of a right pyramid we take 1/3 the area of the base square times the altitude.

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Example 3: The base of a solid is the circular region in the xy-plane bounded by x² + y² = 9.
Find its volume if any cross-section perpendicular to the x-axis is a square.

Solution: The side of each square measures 2y, so the area of a typical square slice or wafer is
A = 4y². However, the slices pile up along the x-axis from – 3 to + 3, so we need it in terms of x.
We know that x² + y² = 9, so we solve for y² to get y² = 9 – x². The area function therefore becomes A(x) = 4( 9 – x² ). Instead of integrating from – 3 to + 3, we'll double the integral from 0 to + 3.

The Volume is

Summary:

1 – Write a formula for the area of a typical slice
2 – Using the base figure's rule of correspondence substitute for the variables,
3 – Integrate the Area formula over the extend of the base figure.

Practice

Draw a sketch for each question and find the volume.

1) A solid's base is the region in the xy-plane bounded by the graphs of y ² = 4x and x = 4.
Find the volume of the solid if every cross section by a plane perpendicular to the x-axis is an isoscele right triangle with one of the equal sides lying in the base of the solid.

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2) A solid's base is the region in the xy-plane bounded by the graph of x ² + y ² = 16. Find the volume of this solid if every cross section by a plane perpendicular to the y-axis is a rectangle with height equal to one third the base.

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3) A solid's base is the region in the xy-plane bounded by the graph of y = 2x² and y = 8.
Find the volume of this solid if every cross section by a plane perpendicular to the x-axis is an equilateral triangle. (Draw a sketch)

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Solutions

1) y ² = 4x and x = 4.

The triangle's base is = its height = 2y
Since area of a triangle is ½bh we get A = ½(4y²) = 2y² = 8x

The volume of the solid is

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2) x ² + y ² = 16 is a circle center (0, 0) with radius = 4.

Since area of a rectangle is bh we get A = (2x)(2/3 x) = 4/3 x²

The volume of the solid is

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3) y = 2x² and y = 8.

The line y = 8 and the curve intersect at x = ± 2

The equilateral triangle's base is 8 - y, so we find the height h using the 30°, 60°, 90° triangle ratios. This gives us that

Since area of a triangle is ½bh we get

The volume of the solid is

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(Caculus II MathRoom Index)

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(all content of the MathRoom Lessons © Tammy the Tutor; 2002 - ).