Antiderivative of the Semicircle: Geometry Meets Calculus
What if you could find the area under a curve not by memorizing a formula, but by drawing a picture? In this lesson, you will discover that the area under a semicircle can be split into a pie-shaped sector and a right triangle – two shapes whose areas you already know. This geometric trick gives you the antiderivative of \(\sqrt{r^2 - x^2}\) without any complicated algebra, and then you will verify it works by differentiating with the product rule and chain rule.
The integral \(\int \sqrt{r^2 - x^2}\,dx\) looks scary, but it connects two beautiful ideas: geometry and calculus. Here is why it matters:
- GPS and navigation: calculating distances along curved paths on Earth’s surface uses integrals involving \(\sqrt{r^2 - x^2}\)
- Engineering: designing arches, tunnels, and domes requires computing areas of circular cross-sections
- Medicine: MRI and CT scanners reconstruct circular slices of your body using these exact integrals
- Astronomy: orbital mechanics involves integrating expressions with \(\sqrt{r^2 - x^2}\) to find swept areas (Kepler’s second law!)
- Computer graphics: rendering smooth curves and circles on screens requires breaking circular regions into tiny triangles and sectors – exactly the decomposition we learn today
Today we discover that by thinking geometrically – splitting a region under a semicircle into a sector and a triangle – we can write down the antiderivative without any algebraic tricks, and then verify it using the product rule and chain rule.
Topics Covered
- The antiderivative \(\int \sqrt{r^2 - x^2}\,dx\) as the area under the upper semicircle
- Decomposing the area into a circular sector plus a right triangle
- Writing the sector area using \(\arccos(x/r)\)
- Writing the triangle area as \(\frac{1}{2}x\sqrt{r^2 - x^2}\)
- The relationship \(\arcsin(x/r) + \arccos(x/r) = \frac{\pi}{2}\)
- Verifying the antiderivative by differentiation using the product rule and chain rule
- Boundary condition: \(A(-r) = 0\) determines the constant of integration
Lecture Video
Key Frames from the Lecture
What You Need to Know First
An antiderivative of a function \(f(x)\) is a function \(F(x)\) whose derivative equals \(f(x)\):
\[F'(x) = f(x)\]
We write this using integral notation:
\[\int f(x)\,dx = F(x) + C\]
The \(C\) is the constant of integration – since the derivative of any constant is zero, there are infinitely many antiderivatives that differ by a constant. We can pin down \(C\) by using a boundary condition (plugging in a known value).
A full circle of radius \(r\) centered at the origin satisfies:
\[x^2 + y^2 = r^2\]
Solving for \(y\) gives two solutions. The upper semicircle is:
\[y = \sqrt{r^2 - x^2}\]
This curve exists only for \(-r \le x \le r\), since we need \(r^2 - x^2 \ge 0\). The area under this curve from \(-r\) to \(r\) is exactly \(\frac{1}{2}\pi r^2\) (half the area of the circle).
The arcsine function \(\arcsin(u)\) answers: “what angle has sine equal to \(u\)?” Similarly, arccosine \(\arccos(u)\) answers: “what angle has cosine equal to \(u\)?”
Key facts:
- \(\arcsin(u)\) returns an angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
- \(\arccos(u)\) returns an angle in \([0, \pi]\)
- They are complementary: \(\arcsin(u) + \arccos(u) = \frac{\pi}{2}\) for all \(|u| \le 1\)
- \(\frac{d}{dx}\arcsin(u) = \frac{1}{\sqrt{1 - u^2}}\) and \(\frac{d}{dx}\arccos(u) = \frac{-1}{\sqrt{1 - u^2}}\)
These derivatives are why arcsine and arccosine show up naturally when integrating expressions involving \(\sqrt{r^2 - x^2}\).
The product rule says that the derivative of a product \(f(x) \cdot g(x)\) is:
\[\frac{d}{dx}[f(x)\cdot g(x)] = f'(x)\cdot g(x) + f(x)\cdot g'(x)\]
You differentiate each factor in turn while keeping the other one fixed, then add the results. It is not \(f'(x) \cdot g'(x)\)!
The chain rule says that the derivative of a composition \(h(g(x))\) is:
\[\frac{d}{dx}[h(g(x))] = h'(g(x)) \cdot g'(x)\]
You take the derivative of the outer function (evaluated at the inner function), then multiply by the derivative of the inner function. For example:
\[\frac{d}{dx}\sqrt{r^2 - x^2} = \frac{1}{2}(r^2 - x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{r^2 - x^2}}\]
The \((-2x)\) comes from the chain rule – it is the derivative of the inner function \(r^2 - x^2\).
A circular sector is a “pie slice” of a circle. If the circle has radius \(r\) and the sector subtends an angle \(\phi\) (in radians), the area is:
\[A_{\text{sector}} = \frac{1}{2}r^2 \phi\]
This formula works because the sector is the fraction \(\frac{\phi}{2\pi}\) of the full circle, and the full circle has area \(\pi r^2\):
\[A_{\text{sector}} = \frac{\phi}{2\pi} \cdot \pi r^2 = \frac{1}{2}r^2 \phi\]
Key Concepts
The Problem: Antiderivative of the Semicircle
We want to find the antiderivative:
\[\int \sqrt{r^2 - x^2}\,dx\]
The function \(y = \sqrt{r^2 - x^2}\) is the upper semicircle of radius \(r\). Geometrically, the antiderivative \(A(x)\) represents the area swept under the curve from \(x = -r\) up to the moving boundary \(x\).
Explore – the upper semicircle \(y = \sqrt{r^2 - x^2}\) with adjustable radius:
Drag the slider \(a\) to sweep the shaded area \(A(a)\) from \(-r\) to \(a\). This swept area is exactly the antiderivative we are computing!
Connecting Area to Antiderivative
As \(x\) advances by a tiny amount \(dx\), the area grows by one thin vertical strip of width \(dx\) and height \(y\):
\[dA = y\,dx = \sqrt{r^2 - x^2}\,dx\]
This tells us the rate of change of the swept area:
\[\frac{dA}{dx} = y = \sqrt{r^2 - x^2}\]
So \(A(x)\) is indeed an antiderivative of \(\sqrt{r^2 - x^2}\). We also have the boundary condition \(A(-r) = 0\), since no area has been swept at the starting point.
The Key Insight: Area = Sector + Triangle
Instead of using algebraic substitution tricks, we decompose the shaded region geometrically. The area under the semicircle from \(-r\) to \(x\) can be split into two parts:
- A circular sector (the “pie slice” from the negative \(x\)-axis to the radius drawn to the point)
- A right triangle (the remaining piece between the radius and the vertical line at \(x\))
Explore – see the sector and triangle decomposition:
The red line is the radius to the point \((a, \sqrt{r^2 - a^2})\). The green lines mark the right triangle. The sector is the pie-shaped region between the negative \(x\)-axis and the radius.
Writing the Sector Area
The sector is bounded by the angle \(\phi\) measured from the negative \(x\)-axis to the radius. We introduce the angle \(\theta\) from the positive \(x\)-axis:
\[\cos\theta = \frac{x}{r} \implies \theta = \arccos\!\left(\frac{x}{r}\right)\]
The sector angle is then:
\[\phi = \pi - \theta = \pi - \arccos\!\left(\frac{x}{r}\right)\]
So the sector area is:
\[A_{\text{sector}} = \frac{1}{2}r^2\phi = \frac{1}{2}r^2\left[\pi - \arccos\!\left(\frac{x}{r}\right)\right]\]
Writing the Triangle Area
The right triangle has:
- Base = \(x\) (the horizontal distance from the origin)
- Height = \(y = \sqrt{r^2 - x^2}\) (the vertical coordinate on the semicircle)
So the triangle area is:
\[A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot \sqrt{r^2 - x^2}\]
The Complete Antiderivative
Adding the two pieces:
The area under a semicircle from \(-r\) to \(x\) splits neatly into a circular sector (a pie slice) and a right triangle. Each piece has a simple formula, and adding them gives the full antiderivative – no algebraic tricks required.
\[\boxed{A(x) = \frac{1}{2}r^2\!\left[\pi - \arccos\!\left(\frac{x}{r}\right)\right] + \frac{1}{2}\,x\sqrt{r^2 - x^2}}\]
Boundary condition check: At \(x = -r\), the swept area should be zero.
- \(\arccos(-r/r) = \arccos(-1) = \pi\), so the sector term gives \(\frac{1}{2}r^2[\pi - \pi] = 0\).
- \(\sqrt{r^2 - r^2} = 0\), so the triangle term gives \(0\).
Indeed \(A(-r) = 0\). Our formula checks out!
Connection to Arcsine
Earlier work showed that part of the antiderivative involves \(\arcsin(x/r)\). Is there a contradiction? No! The identity
The arcsine and arccosine of the same value always add up to \(\frac{\pi}{2}\) (a right angle). This identity lets you freely switch between the two forms of the antiderivative – they differ only by a constant.
\[\arcsin\!\left(\frac{x}{r}\right) + \arccos\!\left(\frac{x}{r}\right) = \frac{\pi}{2}\]
This means \(\arccos(x/r) = \frac{\pi}{2} - \arcsin(x/r)\). Substituting:
\[\pi - \arccos\!\left(\frac{x}{r}\right) = \pi - \frac{\pi}{2} + \arcsin\!\left(\frac{x}{r}\right) = \frac{\pi}{2} + \arcsin\!\left(\frac{x}{r}\right)\]
So we can rewrite the antiderivative equivalently as:
\[A(x) = \frac{1}{2}r^2\!\left[\frac{\pi}{2} + \arcsin\!\left(\frac{x}{r}\right)\right] + \frac{1}{2}\,x\sqrt{r^2 - x^2}\]
The constant \(\frac{1}{2}r^2 \cdot \frac{\pi}{2} = \frac{\pi r^2}{4}\) is just a constant of integration – it vanishes when you differentiate. So both forms are equally valid!
Verifying by Differentiation
To confirm our geometric result, we differentiate \(A(x)\) and check that we recover \(\sqrt{r^2 - x^2}\).
We need:
\[\frac{d}{dx}\!\left[\frac{1}{2}r^2\!\left(\pi - \arccos\!\frac{x}{r}\right) + \frac{1}{2}\,x\sqrt{r^2 - x^2}\right]\]
Term 1 – the sector piece:
\[\frac{d}{dx}\!\left[\frac{1}{2}r^2\!\left(\pi - \arccos\!\frac{x}{r}\right)\right] = \frac{1}{2}r^2 \cdot \frac{1}{\sqrt{1 - x^2/r^2}} \cdot \frac{1}{r}\]
Since \(\frac{d}{dx}\arccos(u) = \frac{-1}{\sqrt{1-u^2}}\) and the minus signs cancel:
\[= \frac{r}{2} \cdot \frac{r}{\sqrt{r^2 - x^2}} = \frac{r^2}{2\sqrt{r^2 - x^2}}\]
Term 2 – the triangle piece, using the product rule on \(\frac{1}{2}x \cdot \sqrt{r^2 - x^2}\):
\[\frac{d}{dx}\!\left[\frac{1}{2}\,x\sqrt{r^2 - x^2}\right] = \frac{1}{2}\!\left[\underbrace{1 \cdot \sqrt{r^2 - x^2}}_{\text{derivative of } x,\;\text{copy } g} + \underbrace{x \cdot \frac{-2x}{2\sqrt{r^2 - x^2}}}_{\text{copy } x,\;\text{chain rule on } g}\right]\]
The chain rule gives us the inner derivative \((-2x)\) of \(r^2 - x^2\):
\[= \frac{1}{2}\!\left[\sqrt{r^2 - x^2} - \frac{x^2}{\sqrt{r^2 - x^2}}\right]\]
Combining over the common denominator \(\sqrt{r^2 - x^2}\):
\[= \frac{1}{2}\cdot\frac{(r^2 - x^2) - x^2}{\sqrt{r^2 - x^2}} = \frac{r^2 - 2x^2}{2\sqrt{r^2 - x^2}}\]
Adding both terms:
\[A'(x) = \frac{r^2}{2\sqrt{r^2 - x^2}} + \frac{r^2 - 2x^2}{2\sqrt{r^2 - x^2}} = \frac{2r^2 - 2x^2}{2\sqrt{r^2 - x^2}}\]
\[= \frac{2(r^2 - x^2)}{2\sqrt{r^2 - x^2}} = \frac{r^2 - x^2}{\sqrt{r^2 - x^2}} = \sqrt{r^2 - x^2}\]
When you differentiate the geometric formula (sector + triangle) using the product rule and chain rule, you get back the semicircle \(\sqrt{r^2 - x^2}\). This confirms that the geometric decomposition is a valid antiderivative – two completely different approaches give the same answer.
It works! The derivative of our geometrically derived \(A(x)\) is indeed \(\sqrt{r^2 - x^2}\). Geometry and calculus agree perfectly.
Why the Constant Does Not Matter
Any two antiderivatives of the same function differ by a constant. Since \(\frac{d}{dx}(\text{constant}) = 0\), the constant disappears when differentiating. This is why both the arcsine version and the arccosine version are equally valid – they differ only by the constant \(\frac{\pi r^2}{4}\).
Cheat Sheet
| What you want | Formula |
|---|---|
| Upper semicircle equation | \(y = \sqrt{r^2 - x^2}\), valid for \(-r \le x \le r\) |
| Antiderivative (arccosine form) | \(\displaystyle\int\!\sqrt{r^2 - x^2}\,dx = \frac{r^2}{2}\!\left[\pi - \arccos\!\frac{x}{r}\right] + \frac{x}{2}\sqrt{r^2 - x^2} + C\) |
| Antiderivative (arcsine form) | \(\displaystyle\int\!\sqrt{r^2 - x^2}\,dx = \frac{r^2}{2}\arcsin\!\frac{x}{r} + \frac{x}{2}\sqrt{r^2 - x^2} + C\) |
| Sector area | \(A_{\text{sector}} = \frac{1}{2}r^2\phi\) where \(\phi\) is the angle in radians |
| Triangle area | \(A_{\text{triangle}} = \frac{1}{2} \cdot \text{base} \cdot \text{height}\) |
| Arcsine-arccosine identity | \(\arcsin(u) + \arccos(u) = \frac{\pi}{2}\) |
| Derivative of \(\arcsin(u)\) | \(\frac{1}{\sqrt{1-u^2}}\) |
| Derivative of \(\arccos(u)\) | \(\frac{-1}{\sqrt{1-u^2}}\) |
| Product rule | \((fg)' = f'g + fg'\) |
| Chain rule | \(\frac{d}{dx}[h(g(x))] = h'(g(x))\cdot g'(x)\) |
The Big Idea
\[\underbrace{\int \sqrt{r^2 - x^2}\,dx}_{\text{calculus problem}} = \underbrace{A_{\text{sector}}}_{\frac{1}{2}r^2(\pi - \arccos\frac{x}{r})} + \underbrace{A_{\text{triangle}}}_{\frac{1}{2}x\sqrt{r^2 - x^2}} \quad \longleftrightarrow \quad \text{geometry solves calculus!}\]