Volumes of Revolution about the x-axis

When a region in the xy-plane is rotated through 2pi radians (360 degrees) about the x-axis, it sweeps out a three-dimensional solid of revolution. Integration allows us to compute the exact volume of such solids using the disc method.

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1 The Disc Method

Derivation

Consider the curve y = f(x) between x = a and x = b, with f(x) >= 0. Divide the interval [a, b] into n thin strips, each of width delta x.

Each strip, when rotated about the x-axis, forms a thin disc (cylinder) with:

• Radius = y = f(x)
• Thickness = delta x

The volume of one disc is approximately:

delta V = pi * y^2 * delta x = pi * [f(x)]^2 * delta x

Summing all discs and taking the limit as delta x -> 0 gives:

V = pi * integral from a to b of y^2 dx = pi * integral from a to b of [f(x)]^2 dx

Critical Warning: You must square y before integrating. A very common error is to compute pi * integral of y dx instead of pi * integral of y^2 dx.

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2 Worked Example 1 — Parabola

Find the volume generated when the region bounded by y = x^2, the x-axis, and the lines x = 0 and x = 3 is rotated completely about the x-axis.

Solution:

V = pi * integral from 0 to 3 of (x^2)^2 dx = pi * integral from 0 to 3 of x^4 dx

= pi * [x^5 / 5] from 0 to 3

= pi * (243/5 - 0)

= 243pi/5 cubic units

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3 Worked Example 2 — Square Root Function

Find the volume when the region between y = sqrt(x), the x-axis, x = 0, and x = 4 is rotated about the x-axis.

Solution:

V = pi * integral from 0 to 4 of (sqrt(x))^2 dx = pi * integral from 0 to 4 of x dx

= pi * [x^2/2] from 0 to 4

= pi * (16/2 - 0)

= 8pi cubic units

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4 Worked Example 3 — Straight Line (Cone)

Find the volume when y = 2x (for 0 <= x <= 3) is rotated about the x-axis.

Solution:

V = pi * integral from 0 to 3 of (2x)^2 dx = pi * integral from 0 to 3 of 4x^2 dx

= pi * [4x^3/3] from 0 to 3

= pi * (36 - 0) = 36pi

This is the volume of a cone with radius 6 and height 3. Checking with the cone formula: V = (1/3) pi r^2 h = (1/3) pi (36)(3) = 36pi. Confirmed!

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5 Worked Example 4 — Trigonometric Function

Find the volume when y = sin x (for 0 <= x <= pi) is rotated about the x-axis.

Solution:

V = pi * integral from 0 to pi of sin^2 x dx

Use the identity sin^2 x = (1 - cos 2x)/2:

= pi * integral from 0 to pi of (1 - cos 2x)/2 dx

= (pi/2) * [x - sin(2x)/2] from 0 to pi

= (pi/2) * [(pi - 0) - (0 - 0)]

= pi^2/2 cubic units

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6 Volume Between Two Curves

When the region between two curves y = f(x) (outer) and y = g(x) (inner) is rotated about the x-axis, the volume is found using the washer method:

V = pi * integral from a to b of ([f(x)]^2 - [g(x)]^2) dx

This subtracts the inner disc from the outer disc at each x-value.

Worked Example 5

Find the volume when the region enclosed between y = x and y = x^2 (from x = 0 to x = 1) is rotated about the x-axis.

Solution:

For 0 <= x <= 1, x >= x^2, so x is the outer curve and x^2 is the inner curve.

V = pi * integral from 0 to 1 of (x^2 - x^4) dx