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6.3 Volumes by Slicing

In single-variable calculus, we find the area under a curve by summing up the areas of thin rectangles. For a function \( y = f(x) \) from \( x = a \) to \( x = b \):

\( \text{area} = \int_{a}^{b} f(x) dx \)

  • \( f(x) \): thin rectangle area at \( x \)
  • \( \int_{a}^{b} \dots dx \): accumulates all from \( x = a \) to \( x = b \)
Graph of y=f(x) showing a thin vertical rectangle of width dx and height f(x) between x=a and x=b.

Volume of this?

To find the volume of a solid, we can use a similar slicing method. If we know the cross-sectional area \( A(x) \) at any point \( x \):

Volume of this slice is area at \( x \) times thickness \( dx \): \( A(x) dx \)

3D solid with a cross-sectional slice highlighted, showing area A(x) and thickness dx.

\( \text{volume} = \int_{a}^{b} A(x) dx \)

\( A(x) \)
area at \( x \)
\( dx \)
thickness of slice
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Example

The base of the object is the region bounded by \( y = 1 - x \) and \( y = -1 + x \) from \( x = 0 \) to \( x = 1 \). Each slice perpendicular to the \( xy \)-plane is a semicircle.

Coordinate graph showing a triangular region bounded by y=1-x and y=-1+x from x=0 to x=1.
A semicircular slice with thickness dx and diameter corresponding to the vertical distance in the base region.
  • Thickness is \( dx \)
  • Half a circle: \( \text{area} = \frac{1}{2} \pi (\text{radius})^2 \)
  • The shaded bar is the diameter of this semicircle

\( \text{diameter} = \text{length of shaded bar} = (1 - x) - (-1 + x) = 2 - 2x \)

\( \text{radius} = 1 - x \)

The volume of each slice = \( \frac{1}{2} \pi (1 - x)^2 dx \)

Stack all possible slices:

\( \int_{0}^{1} \frac{1}{2} \pi (1 - x)^2 dx = \dots = \)

\( \frac{\pi}{6} \)
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Volume of revolution

revolve this region about x-axis
the triangle sweeps out a cone

volume?

A triangle in the first quadrant with vertices at (0,0), (2,0), and (0,2) being revolved around the x-axis to form a cone.

find the volume of each disk at some \(x\), then integrate to accumulate

A coordinate graph showing a line y = 2 - x from x=0 to x=2, with a vertical slice of width dx highlighted.
A single thin disk representing a slice of the cone with radius 2-x and thickness dx.

\(\downarrow\)

\(\text{radius} = \text{height of bar} = 2 - x\)

\(\uparrow y = 0\)

\(\text{volume} = (\text{area})(\text{thickness}) = \pi (\text{radius})^2 dx\)

\[ = \pi (2 - x)^2 dx \]

this is one slice starting at \(x = 0\), ending at \(x = 2\)

integrate to accumulate:

\[ \int_{0}^{2} \pi (2 - x)^2 dx = \dots = \frac{8\pi}{3} \]
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example

volume of solid obtained by revolving the region bounded by \(x = \sqrt{3y}\), \(x = 0\), \(y = 2\), about the y-axis

A graph showing the region bounded by x = sqrt(3y), x = 0, and y = 2 in the first quadrant, with a horizontal slice dy.
\(y = 2\), \(x = 0\), \(y = 0\), \(x = \sqrt{3y}\)
The solid of revolution formed by rotating the region about the y-axis, showing a horizontal disk slice.

\(\text{volume} = \pi (\text{radius})^2 (\text{thickness})\)

\(\uparrow \sqrt{3y} \quad \uparrow dy\)

all these y's and dy \(\rightarrow\) integrate in terms of y

volume of one slice = \(\pi (\sqrt{3y})^2 dy\)

accumulate from \(y = 0\) to \(y = 2\)

\[ \int_{0}^{2} \pi (\sqrt{3y})^2 dy = \int_{0}^{2} 3\pi y dy = 3\pi \left. \frac{y^2}{2} \right|_0^2 = 6\pi \]

this method, accumulate disks (or washers) is called the Disk / Washer Method

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Example: Solid of Revolution

Solid obtained by revolving the region bounded by \( y = \sqrt{x} \), \( y = x \), about the x-axis.

Graph of y=sqrt(x) and y=x on Cartesian axes from x=0 to 1, with a vertical red slice representing a washer.
3D representation of the solid formed by revolving the region, showing a hollow cone space in the center.

hollow cone space

The Washer Method

Diagram of a single washer with outer radius, inner radius, and dx thickness labeled.

washer

Cross-section view of a washer, a red ring with a hole in the middle.

\( \text{area} = \text{area of big} - \text{area of small} \)

\( = \pi(\text{outer})^2 - \pi(\text{inner})^2 \)

Volume Calculation

\( \text{volume of washer} = [\pi(\text{outer})^2 - \pi(\text{inner})^2] dx = (\pi(x) - \pi(x^2)) dx \)

\( \text{outer} = \sqrt{x} \)\( \text{inner} = x \)

Start: \( x = 0 \)

End: \( x = 1 \)

\( \text{volume of whole thing}: \int_{0}^{1} [\pi(x) - \pi(x^2)] dx = \dots = \boxed{\frac{\pi}{6}} \)

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Example: Revolution around \( y = 1 \)

Region is bounded by \( y = 1 \), \( y = x \), \( x = 0 \). Revolved around \( y = 1 \).

Graph showing the triangular region bounded by y=1, y=x, and x=0, with a vertical slice indicated.
3D cone-like solid formed by revolving the triangular region around the line y=1.

\( \text{disk volume} = \pi(\text{radius})^2 \cdot \text{thickness} \)

here, \( dx \) measured from axis of revolution

Detailed graph showing the radius of the disk as the distance between y=1 and y=x, which is 1-x.

\( \int_{0}^{1} \pi(1-x)^2 dx \)

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