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IB Mathematics AI AHL Area and Volume of the region enclosed by a curve MAI Study Notes - New Syllabus

IB Mathematics AI AHL Area and Volume of the region enclosed by a curve MAI Study Notes

LEARNING OBJECTIVE

  • Area of the region enclosed by a curve and the x or y-axes

Key Concepts: 

  • Area of the region enclosed by a curve and the x or y-axes
  • Volumes of revolution about the x- axis or y- axis

MAI HL and SL Notes – All topics

 AREA UNDER A CURVE

Area Under a Curve (x-axis)

The area between a function \( f(x) \) and the x-axis from \( x = a \) to \( x = b \) is calculated using a definite integral:

\( A = \int_a^b f(x)\,dx \)

  • If \( f(x) \geq 0 \), the area is positive.

  • If \( f(x) \leq 0 \), the integral gives a negative value (area below x-axis).

  • If the function crosses the axis, split the integral and use absolute values for total area.

Example
Consider the function: \( f(x) = x^3 – 6x^2 + 8x \)

Find:

  • (a) \( \int_0^2 f(x)\, dx \)
  • (b) \( \int_2^4 f(x)\, dx \)
  • (c) \( \int_0^4 f(x)\, dx \)
  • (d) Total area between the curve and the x-axis within \([0, 4]\)
▶️Answer/Explanation

Solution:

(a) $ \int_0^2 f(x)\, dx = \int_0^2 (x^3 – 6x^2 + 8x)\, dx = \left[\frac{x^4}{4} – 2x^3 + 4x^2\right]_0^2 = 4 – 0 = 4 $

(b) $ \int_2^4 f(x)\, dx = \left[\frac{x^4}{4} – 2x^3 + 4x^2\right]_2^4 = 0 – 4 = -4 $

(c) $ \int_0^4 f(x)\, dx = \left[\frac{x^4}{4} – 2x^3 + 4x^2\right]_0^4 = 0 – 0 = 0 $ [This is the sum of the two earlier integrals]

(d) Total area is: $ A = \int_0^4 |f(x)|\, dx = |4| + |-4| = 8 $

Area Under a Curve (y-axis)

 

When the function is written as \( x = f(y) \), and the region is bounded between \( y = a \) and \( y = b \):

\( A = \int_a^b f(y)\,dy \)

Example

Find the area enclosed between the curve \( x = y^2 \) and the y-axis from \( y = 0 \) to \( y = 2 \).

▶️Answer/Explanation

Since \( x = y^2 \), area with respect to the y-axis is:
$ A = \int_{0}^{2} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3} $
 Final Area: \( \frac{8}{3} \) units².

Area Between Two Curves

To find the area between two curves \( y = f(x) \) (upper) and \( y = g(x) \) (lower) from \( x = a \) to \( x = b \):

\( A = \int_a^b [f(x) – g(x)]\,dx \)

Ensure \( f(x) \geq g(x) \) on the interval \([a, b]\).

  • Use GDC or solve equations to find points of intersection.

  • Use absolute values if functions cross over each other.

Example

Find the area between the curves \( y = x^2 \) and \( y = x+2 \) from \( x = 0 \) to \( x = 2 \).

▶️Answer/Explanation

Solution:

$
\int_{-1}^{2}[(x+2) – x^2]\,dx
= \int_{-1}^{2}(x + 2 – x^2)\,dx
= \left[\frac{x^2}{2} + 2x – \frac{x^3}{3}\right]_{-1}^{2}
$

$
= \left(2 + 4 – \frac{8}{3}\right) – \left(\frac{1}{2} – 2 + \frac{1}{3}\right)
= \frac{10}{3} + \frac{7}{6} = \frac{9}{2}
$

 VOLUMES OF REVOLUTION

 Volumes of Revolution

A solid of revolution is a three-dimensional shape created by spinning a two-dimensional curve around a line within the same plane. The volume of such a solid can be computed using integration. The most typical techniques for determining the volume include the disc method, the shell method, and Washer Method.

 About the x-axis

 

Rotating the graph of \( y = f(x) \) about the x-axis from \( x = a \) to \( x = b \):

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

Example

Find the volume of the solid formed when the region under \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 4 \) is revolved about the x-axis.

▶️Answer/Explanation

Use the formula:
$ V = \pi \int_{0}^{4} (\sqrt{x})^2 \, dx = \pi \int_{0}^{4} x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \cdot \frac{16}{2} = 8\pi $
Volume: \( 8\pi \) units³.

Example
A semicircle of radius \( r \) is defined by the equation \( y = \sqrt{r^2 – x^2} \). This is the upper half of the circle \( x^2 + y^2 = r^2 \).

Find the volume of the solid formed when this semicircle is rotated about the x-axis from \( x = -r \) to \( x = r \).

▶️Answer/Explanation

Solution:

We use the volume of revolution about the x-axis:

$ V = \int_{-r}^{r} \pi y^2 dx $ Since \( y = \sqrt{r^2 – x^2} \),

we have \( y^2 = r^2 – x^2 \),

so: $ V = \int_{-r}^{r} \pi(r^2 – x^2)\, dx $

$ = \pi \left[ r^2x – \frac{x^3}{3} \right]_{-r}^{r} $

$ = \pi \left( \left[r^3 – \frac{r^3}{3}\right] – \left[-r^3 + \frac{r^3}{3} \right] \right) $

$ = \pi \left( \frac{2r^3}{3} + \frac{2r^3}{3} \right) = \frac{4\pi r^3}{3} $

This confirms the standard formula for the volume of a sphere: $ V = \frac{4}{3} \pi r^3 $

 About the y-axis

Rotating \( x = f(y) \) about the y-axis from \( y = c \) to \( y = d \):

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

Example

Find the volume of the solid formed when the region bounded by \( x = y^2 \) and the y-axis from \( y = 0 \) to \( y = 3 \) is revolved about the y-axis.

▶️Answer/Explanation

The radius is \( x = y^2 \). So,
$ V = \pi \int_0^3 (y^2)^2 \, dy = \pi \int_0^3 y^4 \, dy = \pi \left[ \frac{y^5}{5} \right]_0^3 = \pi \cdot \frac{243}{5} $
Final Volume: \( \frac{243\pi}{5} \) units³.

Summary Table

Rotation Axis

Function

Volume Formula

x-axis

\( y = f(x) \)

\( \pi \int_a^b [f(x)]^2 dx \)

y-axis

\( x = f(y) \)

\( \pi \int_c^d [f(y)]^2 dy \)

Technology Tip

  • Use a GDC or graphing software to visualize and calculate integrals and revolved solids.
  • Helpful to check concavity and intersection points when finding enclosed areas.
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