L14: Volume (Area Between Curves)

Lecture on calculating volumes of solids of revolution generated by rotating the region between two curves about the x-axis, y-axis, or a line parallel to one of these axes.

Volume Generated by Area Between Two Curves

When the region bounded by two curves is rotated completely about an axis, the volume is the difference between the outer solid and the inner solid (washer method).

Rotation About the x-axis

For curves $y = g(x)$ (upper/outer) and $y = f(x)$ (lower/inner) on $[a,b]$:

$$V = \pi \int_a^b \left[(g(x))^2 - (f(x))^2\right] dx$$

Equivalently, as the difference of two disk volumes:

$$V = \pi \int_a^b [g(x)]^2 , dx ;-; \pi \int_a^b [f(x)]^2 , dx$$

Rotation About the y-axis

For curves $x = f(y)$ (right/outer) and $x = g(y)$ (left/inner) on $[c,d]$:

$$V = \pi \int_c^d \left[(f(y))^2 - (g(y))^2\right] dy$$

Equivalently:

$$V = \pi \int_c^d [f(y)]^2 , dy ;-; \pi \int_c^d [g(y)]^2 , dy$$

Finding limits: The limits of integration are the x-values (or y-values) where the bounding curves intersect.

Worked Examples

Example 1: $y = x^2$ and $y = x^3$ about the x-axis

The curves intersect where $x^2 = x^3 \Rightarrow x = 0, 1$. On $[0,1]$, $x^2 \ge x^3$.

$$V = \pi \int_0^1 \left[(x^2)^2 - (x^3)^2\right] dx = \pi \int_0^1 \left(x^4 - x^6\right) dx$$

Example 2: $y = \dfrac{4}{x}$, $y = \dfrac{1}{4}$, and $x = 2$ about the x-axis

The region is bounded on the right by $x = 2$ and on the left by the intersection of $y = 4/x$ and $y = 1/4$ at $x = 16$.

  • Bounds: $[2, 16]$
  • Outside radius: $\dfrac{4}{x}$
  • Inside radius: $\dfrac{1}{4}$

$$V = \int_2^{16} \left[\pi\left(\frac{4}{x}\right)^2 - \pi\left(\frac{1}{4}\right)^2\right] dx$$

Example 3: $y = 2x^2$ and $y = 3x - 1$ about the x-axis

Intersection points: $2x^2 = 3x - 1 \Rightarrow 2x^2 - 3x + 1 = 0 \Rightarrow x = \dfrac{1}{2}, 1$.

  • Bounds: $\left[\dfrac{1}{2}, 1\right]$
  • Outside radius: $3x - 1$
  • Inside radius: $2x^2$

$$V = \int_{1/2}^{1} \left[\pi(3x-1)^2 - \pi(2x^2)^2\right] dx$$

Example 4: $y = \sqrt[3]{x}$ and $y = \dfrac{x}{4}$ in the first quadrant about the y-axis

The region in the first quadrant bounded by $y = \sqrt[3]{x}$ and $y = x/4$ is rotated about the y-axis.

Rewriting: $x = y^3$ and $x = 4y$. The curves intersect at $(0,0)$ and $(8,2)$.

  • Bounds (in $y$): $[0, 2]$
  • Right (outer) function: $x = 4y$
  • Left (inner) function: $x = y^3$

$$V = \pi \int_0^2 \left[(4y)^2 - (y^3)^2\right] dy = \pi \int_0^2 \left(16y^2 - y^6\right) dy$$

Volume About Lines Parallel to the Axes

When the axis of rotation is a horizontal line $y = k$ (parallel to the x-axis), the radius of each curve is its vertical distance from $y = k$. Similarly, for a vertical line $x = h$, the radius is the horizontal distance from $x = h$.

Key principle: Radius = (curve value) $-$ (axis value). When the axis lies below the region, subtracting a negative number increases the radius.

Example: Region bounded by $y = 2 - x^2$ and $y = 1$

The curves intersect where $2 - x^2 = 1 \Rightarrow x = \pm 1$. Bounds are always $[-1, 1]$.

(a) About the line $y = 1$

Since $y = 1$ is one of the bounding curves, this reduces to the disk method.

  • Radius: $(2 - x^2) - 1 = 1 - x^2$

$$V = \int_{-1}^{1} \pi(1 - x^2)^2 , dx$$

(b) About the x-axis ($y = 0$)

Now $y = 1$ creates a hole, so we use the washer method.

  • Outside radius: $2 - x^2$
  • Inside radius: $1$

$$V = \int_{-1}^{1} \left[\pi(2-x^2)^2 - \pi(1)^2\right] dx$$

(c) About the line $y = -1$

Both curves are above $y = -1$, so both radii are shifted upward by 1.

  • Outside radius: $(2 - x^2) - (-1) = 3 - x^2$
  • Inside radius: $1 - (-1) = 2$

$$V = \int_{-1}^{1} \left[\pi(3-x^2)^2 - \pi(2)^2\right] dx$$

Summary Table: Choosing the Approach

Axis of Rotation Method Radius Formula
x-axis ($y = 0$) Washer (if hole exists) $R(x) = g(x)$, $r(x) = f(x)$
y-axis ($x = 0$) Washer (if hole exists) $R(y) = f(y)$, $r(y) = g(y)$
Horizontal line $y = k$ Washer/Disk Vertical distance: $(y_{\text{curve}} - k)$
Vertical line $x = h$ Washer/Disk Horizontal distance: $(x_{\text{curve}} - h)$

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