- #1

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- Homework Statement
- Calculate the area of the segment of the circle shown in the figure below, shaded in red.

- Relevant Equations
- 1. The area of the sector of a circle that subtends an angle of ##\theta^{\circ}## at the center is ##A = \dfrac{\theta^{\circ}}{360^{\circ}}\times\pi r^2 ##

2. The cosine of an angle ##\theta## in a right angled triangle : ##\cos\theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}##

3. Area of a triangle : ##A = \dfrac{1}{2}\times\; \text{base}\; \times\; \text{height}##.

4. The equation of a circle of radius ##a## : ##y(x) = \sqrt{a^2-x^2}##.

5. The area that a function ##f(x)## subtends with the horizontal from ##x_1\rightarrow x_2## is given by ##A = \int_{x_1}^{x_2} f(x) dx##.

5. ##\int\sqrt{a^2-x^2} = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}##

**To find the area of the shaded segment filled in
Problem Statement : **

**red**in the circle shown to the right. The region is marked by the points

**PQRP**.

**
**

**Attempt 1 (**I mark some relevant lengths inside the circle, shown left. Clearly

__without__calculus):**OS = 9 cm**and

**SP = 12 cm**using the Pythagorean theorem. The angle

**##\theta = \angle \mathbf{SOP}##**(##=\angle\mathbf{SOR}##) can be found : ##\theta=\cos^{-1}\frac{9}{15}\approx 53^{\circ}##. Hence the total angle

**ROP**: ##2\theta \approx 106^{\circ}##. The area of the sector

**ORQPO**: ##A_{\text{sect}}= \dfrac{106^{\circ}}{360^{\circ}}\times \pi \times 15^2 = 208.1\;\text{cm}^2##. The area of the full triangle

**ROP**: ##A_{\text{tri}} = \dfrac{1}{2}\times\;\text{24 cm}\;\times\;\text{9 cm} = 108\;\text{cm}^2##. Hence the area of the segment

**RSPQR**shaded in

**red**: ##A_{\text{seg}}= 208.1-108\approx \boxed{100\;\text{cm}^2}##.

**The area of the rectangle
Attempt 2 (using calculus) : **

**PTQS**is ##A_{\text{rect}} = 12\times 6 = 72\;\text{cm}^2##. What is the area of figure

**PQSP**shaded in

**green**? The equation of the arc

**QP**is known to be ##y=(15^2-x^2)^{\frac{1}{2}}##. The area that this arc subtends with the horizontal can be found calculated :

$$A_{\text{arc}}= \int_0^{12} \sqrt{15^2-x^2}dx= \left|\dfrac{x}{2}\sqrt{15^2-x^2}\right|_0^{12}+\left|\dfrac{15^2}{2}\sin^{-1}\dfrac{x}{15}\right|_0^{12}$$

$$ A_{\text{arc}} = 6\sqrt{15^2-12^2}+\dfrac{15^2}{2}\times\underbrace{\sin^{-1}\dfrac{4}{5}}_{\text{to be evaluated in radians}}$$

$$A_{\text{arc}} = 158.32\;\text{cm}^2\;!$$

The area under the arc cannot be more than the area of the (larger) rectangle = 72 cm^2. Where am I mistaken in calculating my integral?

**A hint or suggestion is welcome.**