Grade 12

Advanced

10 min

Lesson 7 of 12

Applications of Integration: Area and Volume

Not Started

Utilize integration to calculate the area between curves and the volume of solids of revolution.

Have you ever wondered how a 3D printer knows exactly how much material is needed for a curved object, or how engineers calculate the weight of a massive dome? The secret lies in using calculus to 'slice' the world into infinite pieces and add them back together.

Learning Objectives

1.Calculate the area of a region bounded by two or more intersecting curves using definite integrals.
2.Determine the volume of a solid of revolution using the Disk and Washer methods.
3.Set up and evaluate integrals for physical applications such as work and fluid pressure.

The Area Between Curves

To find the area between two curves,

f(x)

and

g(x)

, we subtract the 'lower' function from the 'upper' function over a specific interval

[a, b]

. This is because the integral

\int_a^b f(x) dx

represents the area under

f(x)

down to the x-axis. By subtracting

\int_a^b g(x) dx

, we effectively remove the unwanted space below the bottom curve. The general formula is:

A = \int_a^b [f(x)_{top} - g(x)_{bottom}] dx

If the curves intersect, you must first find the points of intersection by setting

f(x) = g(x)

to determine your limits of integration

a

and

b

Find the area of the region bounded by $f(x) = x^2$ and $g(x) = x$ .

1. Find intersections:

x^2 = x \implies x^2 - x = 0 \implies x(x-1) = 0

. So,

x=0

and

x=1

. 2. Identify the top function: On

[0, 1]

x \ge x^2

, so

g(x) = x

is the top. 3. Set up the integral:

A = \int_0^1 (x - x^2) dx

4. Evaluate:

[\frac{1}{2}x^2 - \frac{1}{3}x^3]_0^1 = (\frac{1}{2} - \frac{1}{3}) - 0 = \frac{1}{6}

Quick Check

If you are finding the area between $y=7$ and $y=2$ from $x=0$ to $x=5$ , what is the resulting area?

Subtract the bottom from the top (7 - 2) and integrate over the interval 0 to 5.

Answer

Solids of Revolution: The Disk Method

Imagine taking a 2D shape and spinning it rapidly around an axis (like the x-axis). This creates a solid of revolution. To find its volume, we slice the solid into thin circular disks. Each disk has a thickness of

dx

and a radius

R(x)

equal to the function's height. Since the area of a circle is

\pi r^2

, the volume of one thin disk is

\pi [R(x)]^2 dx

. Summing these up gives the Disk Method formula:

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

Find the volume of the solid formed by rotating $y = \sqrt{x}$ about the x-axis from $x=0$ to $x=4$ .

1. Identify the radius:

R(x) = \sqrt{x}

. 2. Set up the integral:

V = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx

3. Evaluate:

\pi [\frac{1}{2}x^2]_0^4 = \pi (\frac{1}{2}(16) - 0) = 8\pi

Quick Check

In the Disk Method, what geometric shape represents the cross-section of the solid?

Answer

A circle (or disk).

The Washer Method and Physical Work

When the region being rotated does not touch the axis of revolution, it creates a hole in the center, forming a washer. The volume is the outer volume minus the inner volume:

V = \pi \int_a^b ([R_{outer}(x)]^2 - [r_{inner}(x)]^2) dx

Beyond geometry, integration applies to Work. Work is defined as force times displacement. If the force

F(x)

varies, we use:

W = \int_a^b F(x) dx

This is commonly used in Hooke's Law for springs (

F=kx

) or pumping liquids out of tanks.

A force of 40N is required to hold a spring that has been stretched from its natural length of 10cm to 15cm. How much work is done stretching it from 15cm to 18cm?

1. Find the spring constant

k

F = kx

40 = k(0.05m) \implies k = 800 N/m

. 2. Define the limits: From 15cm to 18cm is a displacement from

x=0.05

x=0.08

meters. 3. Set up the integral:

W = \int_{0.05}^{0.08} 800x dx

4. Evaluate:

[400x^2]_{0.05}^{0.08} = 400(0.0064 - 0.0025) = 400(0.0039) = 1.56 Joules

Key Takeaways

1Area between curves is found by integrating the difference between the upper and lower functions: $\int (f_{top} - g_{bottom}) dx$ .
2The Disk Method calculates volume by summing circular cross-sections: $V = \pi \int [R(x)]^2 dx$ .
3The Washer Method accounts for hollow centers by subtracting the inner radius squared from the outer radius squared.
4Physical Work is the integral of a variable force function over a distance interval.

Quick Check

Multiple Choice

Which formula is used to find the volume of a solid with a hollow center rotated around the x-axis?

Multiple Choice

To find the area between $y = x^2 + 2$ and $y = x$ from $x=0$ to $x=2$ , which function is the 'top' function?

True/False

The Disk Method requires multiplying the integral by $\pi$ because the cross-sections are circles.

What's Next?

Review Tomorrow

In 24 hours, try to write down the formulas for the Disk Method and the Washer Method from memory and explain the difference between them.

Practice Activity

Find a household object with circular symmetry (like a soda can or a funnel) and try to sketch the 2D function that would create it if rotated around an axis.

Browse

Browse

Applications of Integration: Area and Volume

Learning Objectives

The Area Between Curves

Solids of Revolution: The Disk Method

The Washer Method and Physical Work

Key Takeaways

Quick Check

What's Next?

Applications of Integration: Area and Volume

Learning Objectives

The Area Between Curves

Solids of Revolution: The Disk Method

The Washer Method and Physical Work

Key Takeaways

Quick Check

What's Next?