M2 Chapter 11: Applications of Definite Integration

Hey everyone! Welcome to one of the most visual and satisfying topics in M2 Calculus. Ever wondered how architects calculate the volume of a curved dome, or how animators find the area of a complex shape in a design? The secret is definite integration!

In this chapter, we're moving beyond the abstract formulas and using our integration skills to solve real-world geometry problems. We'll learn how to find the exact area of curvy shapes and the precise volume of 3D objects. It's where the numbers on the page turn into shapes you can actually imagine.

Don't worry if integration still seems a bit tricky. We'll break everything down step-by-step. Let's get started!


Part 1: Finding the Area of a Plane Figure

The Big Idea: A Quick Recap

Remember that a definite integral, $$ \int_a^b f(x) \,dx $$, gives us the signed area between the curve $$y = f(x)$$ and the x-axis, from $$x=a$$ to $$x=b$$.

Analogy Time! Imagine you're slicing a weirdly shaped loaf of bread. The integral is like a machine that slices the area under the curve into infinitely many, super-thin vertical strips (rectangles), calculates the area of each tiny strip, and then adds them all up to get the total area.

The key word here is "signed" area.

  • If the curve is above the x-axis, the area is positive.
  • If the curve is below the x-axis, the area is negative.
This is a super important point, so keep it in mind!


Area Bounded by a Curve and the x-axis

This is the most direct application. We want to find the physical area, which must always be a positive value.

Step-by-Step Guide:
  1. Sketch the graph: This is the most important step! A quick sketch helps you see if the area is above or below the x-axis.
  2. Find the boundaries: Determine the limits of integration, a and b. These are either given in the question or are the x-intercepts of the curve.
  3. Set up the integral:
    • If the area is entirely above the x-axis ($$f(x) \ge 0$$), the area is simply $$ \text{Area} = \int_a^b f(x) \,dx $$.
    • If the area is entirely below the x-axis ($$f(x) \le 0$$), the integral will be negative. To get a positive area, we use $$ \text{Area} = -\int_a^b f(x) \,dx $$ or $$ \text{Area} = \left| \int_a^b f(x) \,dx \right| $$.
  4. What if it crosses? If the curve crosses the x-axis (partly above, partly below), you must split the integral into separate parts and make sure each part gives a positive area.
Example: Area below the x-axis

Find the area of the region enclosed by the curve $$y = x^2 - 4$$, the x-axis, and the lines $$x=0$$ and $$x=2$$.

  1. Sketch: We know $$y=x^2-4$$ is a parabola opening upwards, shifted down by 4. Between x=0 and x=2, the curve is below the x-axis.
  2. Boundaries: Given as $$a=0$$ and $$b=2$$.
  3. Set up: Since the area is below the x-axis, we need to add a negative sign to make the result positive. $$ \text{Area} = -\int_0^2 (x^2 - 4) \,dx $$
  4. Integrate: $$ \text{Area} = -\left[ \frac{x^3}{3} - 4x \right]_0^2 $$ $$ \text{Area} = -\left( \left( \frac{2^3}{3} - 4(2) \right) - \left( \frac{0^3}{3} - 4(0) \right) \right) $$ $$ \text{Area} = -\left( \frac{8}{3} - 8 \right) = -\left( -\frac{16}{3} \right) = \frac{16}{3} \text{ sq. units} $$
Common Mistakes to Avoid:

If a curve crosses the x-axis, say at $$x=c$$, do NOT just integrate from $$a$$ to $$b$$. You must calculate $$ \int_a^c f(x) \,dx $$ and $$ \int_c^b f(x) \,dx $$ separately, take the absolute value of each, and then add them. Integrating straight through will cause the positive and negative areas to cancel each other out, giving you the wrong answer!


Area Between Two Curves

What if we want the area trapped between two functions, say $$y=f(x)$$ and $$y=g(x)$$?

Analogy Time! Imagine you have a big rectangular piece of cardboard ($$\int f(x)dx$$) and you cut a smaller, curvy shape out of it ($$\int g(x)dx$$). The area of the leftover piece is the area of the big rectangle minus the area of the cutout.

It's the same idea here! The formula is:

Area between curves = $$ \int_a^b (\text{Top curve} - \text{Bottom curve}) \,dx $$

Step-by-Step Guide:
  1. Sketch the graphs: Draw both curves to see which one is on top and which is on the bottom.
  2. Find intersections: To find the limits of integration, a and b, set the two functions equal to each other ($$f(x) = g(x)$$) and solve for x.
  3. Identify Top/Bottom: In the interval $$[a, b]$$, determine which function has the larger y-value (the top curve).
  4. Set up and integrate: Use the formula $$ \text{Area} = \int_a^b [f(x)_{\text{top}} - g(x)_{\text{bottom}}] \,dx $$.
Memory Aid:

Just remember "Top minus Bottom". This simple rule works even if both curves are below the x-axis, because the subtraction automatically handles the signs for you!

Example: Area between two curves

Find the area of the region enclosed by $$y=x+2$$ and $$y=x^2$$.

  1. Sketch: $$y=x+2$$ is a straight line. $$y=x^2$$ is a parabola. The line is above the parabola in the enclosed region.
  2. Intersections: Set them equal. $$ x^2 = x+2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 $$. So they intersect at $$x=-1$$ and $$x=2$$. These are our limits!
  3. Top/Bottom: Between -1 and 2, the line $$y=x+2$$ is the top curve, and $$y=x^2$$ is the bottom curve.
  4. Integrate: $$ \text{Area} = \int_{-1}^2 [(\text{Top}) - (\text{Bottom})] \,dx = \int_{-1}^2 [(x+2) - (x^2)] \,dx $$ $$ \text{Area} = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^2 $$ $$ \text{Area} = \left( \frac{2^2}{2} + 2(2) - \frac{2^3}{3} \right) - \left( \frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} \right) $$ $$ \text{Area} = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) = \left( \frac{10}{3} \right) - \left( -\frac{7}{6} \right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} \text{ sq. units} $$
A Quick Note on Integrating with respect to y

Sometimes, it's easier to slice the area horizontally. In this case, our functions are in the form $$x = f(y)$$ and the formula becomes:

Area = $$ \int_c^d (\text{Right curve} - \text{Left curve}) \,dy $$

This is less common, but can be a huge shortcut for certain problems!

Key Takeaway for Area

Finding area is all about adding up infinite tiny strips. Always sketch the graph first! For area between a curve and the x-axis, watch out for negative regions. For area between two curves, just remember Top - Bottom or Right - Left.



Part 2: Finding the Volume of a Solid of Revolution

The Big Idea: From 2D Area to 3D Volume

Now for the really cool part! What happens if you take a 2D area and spin it around an axis? You get a 3D object called a solid of revolution.

Analogy Time! Think of a potter's wheel. A potter starts with a 2D profile (a curve) and spins it to create a 3D pot. That's exactly what we are doing with mathematics!

  • Spin a semi-circle around its diameter, and you get a sphere.
  • Spin a rectangle around one of its sides, and you get a cylinder.
  • Spin a triangle around one of its legs, and you get a cone.

The Disc Method

The syllabus requires us to use the Disc Method. It's a beautifully simple idea. Remember those thin vertical rectangles we used for area? If we spin one of those rectangles around the x-axis, it sweeps out the shape of a thin disc or a coin.

  • The radius (r) of this disc is the height of the rectangle, which is just the function's value, $$y = f(x)$$.
  • The thickness of this disc is a tiny change in x, which we call $$dx$$.
  • The volume of a single disc is (Area of face) × (thickness) = $$ (\pi r^2) \times dx = \pi [f(x)]^2 \,dx $$.

To find the total volume, we just add up (integrate) the volumes of all these infinite discs!

Volume (rotation about x-axis) = $$ V = \pi \int_a^b [f(x)]^2 \,dx $$

Common Mistakes to Avoid:

There are two classic mistakes everyone makes at least once:

  1. Forgetting to SQUARE the function! It's $$ \pi r^2 $$, so you must square the function inside the integral.
  2. Forgetting the $$ \pi $$! Don't leave $$ \pi $$ behind. It's a crucial part of the volume formula for a cylinder.

Step-by-Step Guide (Rotation about x-axis):
  1. Sketch the 2D region you are rotating.
  2. Identify the axis of rotation (e.g., the x-axis).
  3. Determine the limits of integration, a and b.
  4. Find the radius, $$r(x)$$. For rotation about the x-axis, the radius is simply $$r(x) = y = f(x)$$.
  5. Set up and integrate using the formula $$ V = \pi \int_a^b [r(x)]^2 \,dx $$.
Example: Disc Method

Find the volume of the solid generated by revolving the region under $$y=\sqrt{x}$$ from $$x=1$$ to $$x=4$$ about the x-axis.

  1. Sketch: The curve $$y=\sqrt{x}$$ starts at the origin and curves upwards.
  2. Axis: x-axis.
  3. Limits: Given as $$a=1, b=4$$.
  4. Radius: The radius is the distance from the x-axis up to the curve, so $$r(x) = \sqrt{x}$$.
  5. Integrate: $$ V = \pi \int_1^4 (\sqrt{x})^2 \,dx = \pi \int_1^4 x \,dx $$ $$ V = \pi \left[ \frac{x^2}{2} \right]_1^4 $$ $$ V = \pi \left( \frac{4^2}{2} - \frac{1^2}{2} \right) = \pi \left( \frac{16}{2} - \frac{1}{2} \right) = \frac{15\pi}{2} \text{ cubic units} $$

The Washer Method (A Hole in the Middle)

What if we rotate the area between two curves? We get a solid with a hole in it, like a washer, a donut, or a lampshade. The logic is simple: Volume of Big Solid - Volume of Hole.

  • Let $$R(x)$$ be the outer radius (from the axis of rotation to the outer curve).
  • Let $$r(x)$$ be the inner radius (from the axis of rotation to the inner curve).

The volume formula is:

Volume (Washer Method) = $$ V = \pi \int_a^b ( [R(x)]^2 - [r(x)]^2 ) \,dx $$

This is just applying the disc method twice and subtracting!


Rotation About Other Axes

The syllabus also requires rotation about lines parallel to the axes, like $$y=k$$ or $$x=h$$. Don't panic! The method is the same. The only thing that changes is how you define the radius.

The key is to always define the radius as the distance from the axis of rotation to the curve.

Let's say we rotate the region under $$y=f(x)$$ around the horizontal line $$y=k$$:

  • The radius is the vertical distance between the curve and the line: $$ \text{radius} = |f(x) - k| $$.
  • To find this distance, you always do (Top - Bottom). So, if the curve is above the line, radius = $$f(x)-k$$. If the line is above the curve, radius = $$k-f(x)$$.
  • The volume formula becomes: $$ V = \pi \int_a^b (\text{radius})^2 \,dx = \pi \int_a^b (f(x)-k)^2 \,dx $$.

The same logic applies for rotating around a vertical line $$x=h$$, but you'll be integrating with respect to y. The radius will be $$|g(y)-h|$$, calculated as (Right - Left).

Did you know?

The Greek mathematician Archimedes used a method incredibly similar to this over 2000 years ago! He mentally sliced up shapes into thin pieces to find their volumes and areas, a method called the "method of exhaustion." He was doing calculus without even calling it calculus!

Key Takeaway for Volume

Volume of revolution is found by adding up infinite thin discs. The master formula is $$ V = \pi \int (\text{radius})^2 \times (\text{thickness}) $$. The hardest part is correctly identifying the radius based on the axis of rotation. Always draw a picture to help you see it!


Final Summary & Quick Reference

Area
  • Area under $$f(x)$$ vs. x-axis: $$ \int_a^b f(x) \,dx $$. (Be careful with regions below the axis!)
  • Area between $$f(x)$$ and $$g(x)$$: $$ \int_a^b (\text{Top Curve} - \text{Bottom Curve}) \,dx $$.
Volume (Disc/Washer Method)
  • Rotation about x-axis:
    Radius $$r(x) = f(x)$$.
    $$ V = \pi \int_a^b [f(x)]^2 \,dx $$
  • Rotation about y-axis:
    Function must be $$x=g(y)$$. Radius $$r(y) = g(y)$$.
    $$ V = \pi \int_c^d [g(y)]^2 \,dy $$
  • Rotation about horizontal line $$y=k$$:
    Radius $$r(x) = |f(x)-k|$$.
    $$ V = \pi \int_a^b [f(x)-k]^2 \,dx $$
  • For regions with holes (Washer Method):
    Use $$ V = \pi \int_a^b ([\text{Outer Radius}]^2 - [\text{Inner Radius}]^2) \,dx $$.

You've reached the end! These applications are a big step, but they are also very systematic. The key to success is to always, always, always draw a sketch of the region and a representative rectangle/disc. This will help you visualize the problem and set up the correct integral.

Keep practicing, and you'll master this in no time. You've got this!