IB Mathematics AA AHL Area of the region enclosed by a curve and y axis Study Notes - New Syllabus

IB Mathematics AA AHL Area of the region enclosed by a curve and y axis Study Notes

LEARNING OBJECTIVE

Area of the region

Key Concepts:

Area of the region
Volumes of revolution

Area of the Region Enclosed by a Curve and the y-axis

The area of a region bounded by a curve and the y-axis is a common problem in integral calculus. The approach depends on whether the curve is given in the form $ y = f(x) $ or $ x = g(y) $, and the interval over which the area is considered.

1. When the curve is given as y = f(x)

Suppose the curve is described by $ y = f(x) $, and the region is enclosed between the curve, the y-axis $ (x=0) $, and the vertical lines $ x = a $ and $ x = b $, with $ 0 \leq a < b $.

$ \boxed{\text{Area} = \int_{a}^{b} |f(x)| \, dx} $

The absolute value is important because the area is always positive, even if $ f(x) $ is negative somewhere in the interval. If $ f(x) \geq 0 $ throughout $ [a, b] $, you can omit the absolute value:

$\boxed{ \text{Area} = \int_{a}^{b} f(x) \, dx} $

Steps to find the area:

Identify the function $ f(x) $ and the interval $ [a, b] $.
Determine if $ f(x) $ changes sign on $ [a, b] $. If yes, split the integral at points where $ f(x) = 0 $.
Set up and evaluate the definite integral $ \int_a^b |f(x)| \, dx $.

2.When the curve is given as x = g(y)

In some cases, the curve is better expressed as $ x = g(y) $. If the region is bounded by the curve, the y-axis $ (x = 0) $, and horizontal lines $ y = c $ and $ y = d $, where $ c < d $, the area is found by integrating with respect to $ y $:

$ \boxed{\text{Area} = \int_{c}^{d} g(y) \, dy }$

This integral sums the horizontal slices between the y-axis and the curve over the vertical interval.

Steps to find the area:

Express the curve as $ x = g(y) $.
Identify the limits $ y = c $ and $ y = d $ over which to integrate.
Set up the definite integral $ \int_c^d g(y) \, dy $.
Evaluate the integral to find the area.

Note:

If the function crosses the axes, split the integral accordingly to ensure positive area contributions.
Always sketch the region and identify all boundaries before integrating.
Check the sign of $ f(x) $ or $ g(y) $ to decide if absolute values are necessary.

Example 1

Calculate the area enclosed by the curve $ y = x^2 $, the y-axis, and the line $ x = 3 $.

▶️ Answer/Explanation

Solution:

The function is $ y = x^2 $ and the interval is $ [0,3] $.
Since $ y = x^2 \geq 0 $ for $ x \geq 0 $, we can integrate directly.
Setup the integral:
$ \text{Area} = \int_0^3 x^2 \, dx $
Evaluate the integral:
$ \int_0^3 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^3 = \frac{27}{3} – 0 = 9 $
The area enclosed is $ \boxed{9} $ square units.

Example 2

Find the area enclosed by the curve $ x = y^2 $, the y-axis $ (x=0) $, and the line $ y = 2 $.

▶️ Answer/Explanation

Solution:

The curve is $ x = g(y) = y^2 $.
The limits for $ y $ are from 0 to 2.
Setup the integral:
$ \text{Area} = \int_0^2 y^2 \, dy $
Evaluate the integral:
$ \int_0^2 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3} – 0 = \frac{8}{3} $
The area enclosed is $ \boxed{\frac{8}{3}} $ square units.

Volumes of Revolution about the x-axis

The volume of a solid formed by revolving a region bounded by a curve and the x-axis around the x-axis can be calculated using integral calculus. This method applies when the region lies between the curve $ y = f(x) $, the x-axis $ y = 0 $, and vertical lines $ x = a $ and $ x = b $.

Formula:

$ \boxed{\text{Volume} = \pi \int_a^b [f(x)]^2 \, dx }$

This integral sums up the volumes of infinitesimally thin disks (or washers) perpendicular to the x-axis, each having radius $ f(x) $ and thickness $ dx $.

Steps to find the volume:

Identify the curve $ y = f(x) $ and the interval $ [a, b] $ over which the region is revolved.
Express the volume formula using $ \pi \int_a^b [f(x)]^2 \, dx $.
Evaluate the definite integral to find the volume.
If necessary, use algebraic simplification or substitution before integrating.

Note:

If the curve dips below the x-axis, consider the square of $ f(x) $ to ensure radius is positive (since radius is a distance).
If the solid has a hole (washer method), subtract the volume generated by the inner curve.
Always sketch the region and solid to understand the shape before integration.

Example 1

Find the volume of the solid formed by revolving the region bounded by $ y = \sqrt{x} $, the x-axis, and the lines $ x=0 $ and $ x=4 $ about the x-axis.

▶️ Answer/Explanation

Solution:

The curve is $ y = \sqrt{x} $, and the interval is $ [0,4] $.
Volume formula:
$ V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx $
Evaluate the integral:
$ \pi \int_0^4 x \, dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \left( \frac{16}{2} – 0 \right) = 8\pi $
The volume of the solid is $ \boxed{8\pi} $ cubic units.

Example 2

Calculate the volume of the solid obtained by rotating the region bounded by $ y = x^2 $, the x-axis, and $ x = 1 $ about the x-axis.

▶️ Answer/Explanation

Solution:

The curve is $ y = x^2 $ on the interval $ [0,1] $.
Volume formula:
$ V = \pi \int_0^1 (x^2)^2 \, dx = \pi \int_0^1 x^4 \, dx $
Evaluate the integral:
$ \pi \int_0^1 x^4 \, dx = \pi \left[ \frac{x^5}{5} \right]_0^1 = \frac{\pi}{5} $
The volume of the solid is $ \boxed{\frac{\pi}{5}} $ cubic units.

Volumes of Revolution about the y-axis

The volume of a solid formed by revolving a region bounded by a curve and the y-axis around the y-axis can be calculated using integral calculus. This method applies when the region lies between the curve $ x = g(y) $, the y-axis $ x = 0 $, and horizontal lines $ y = c $ and $ y = d $.

Formula:

$ \boxed{\text{Volume} = \pi \int_c^d [g(y)]^2 \, dy} $

This integral sums the volumes of infinitesimally thin disks perpendicular to the y-axis, each having radius $ g(y) $ and thickness $ dy $.

Steps to find the volume:

Identify the curve $ x = g(y) $ and the interval $ [c, d] $ over which the region is revolved.
Set up the volume formula using $ \pi \int_c^d [g(y)]^2 \, dy $.
Evaluate the definite integral to find the volume.
Use algebraic simplification or substitution if necessary before integrating.

Note:

If the curve crosses the y-axis or dips below it, ensure the radius $ g(y) $ is positive, since it represents a distance.
If the solid has a hole (washer method), subtract the inner volume accordingly.
Sketching the region and solid helps visualize the problem and identify limits correctly.

Example 1

Find the volume of the solid formed by revolving the region bounded by $ x = y^2 $, the y-axis $ (x=0) $, and the lines $ y=0 $ and $ y=3 $ about the y-axis.

▶️ Answer/Explanation

Solution:

The curve is $ x = y^2 $ with limits $ y = 0 $ to $ y = 3 $.
Volume formula:
$ V = \pi \int_0^3 (y^2)^2 \, dy = \pi \int_0^3 y^4 \, dy $
Evaluate the integral:
$ \pi \int_0^3 y^4 \, dy = \pi \left[ \frac{y^5}{5} \right]_0^3 = \pi \left( \frac{243}{5} – 0 \right) = \frac{243\pi}{5} $
The volume of the solid is $ \boxed{\frac{243\pi}{5}} $ cubic units.

Example 2

Calculate the volume of the solid obtained by rotating the region bounded by $ x = 2y $, the y-axis, and $ y = 4 $ about the y-axis.

▶️ Answer/Explanation

Solution:

The curve is $ x = 2y $ on the interval $ y \in [0,4] $.
Volume formula:
$ V = \pi \int_0^4 (2y)^2 \, dy = \pi \int_0^4 4y^2 \, dy $
Evaluate the integral:
$ \pi \int_0^4 4y^2 \, dy = 4\pi \left[ \frac{y^3}{3} \right]_0^4 = 4\pi \left( \frac{64}{3} \right) = \frac{256\pi}{3} $
The volume of the solid is $ \boxed{\frac{256\pi}{3}} $ cubic units.

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IB Mathematics AA AHL Area of the region enclosed by a curve and y axis Study Notes

IB Mathematics AA AHL Area of the region enclosed by a curve and y axis Study Notes - New Syllabus

IB Mathematics AA AHL Area of the region enclosed by a curve and y axis Study Notes

Area of the Region Enclosed by a Curve and the y-axis

Volumes of Revolution about the x-axis

Volumes of Revolution about the y-axis

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