Finding Area of Shaded Segment in Circle Using Calculus

[brotherbobby]

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

Problem Statement : To find the area of the shaded segment filled in red in the circle shown to the right. The region is marked by the points PQRP.

​

Attempt 1 (without calculus): I mark some relevant lengths inside the circle, shown left. Clearly 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 : $A_{\text{seg}}= 208.1-108\approx \boxed{100\;\text{cm}^2}$.

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

 

[anuttarasammyak]

$\triangle POR= 12*9$
OPQR as a part of the circle =$15^2 *2 \arctan \frac{4}{3}$
Red area = $15^2 *2 \arctan \frac{4}{3} - 12*9$

 

[Orodruin]

The equation of the arc QP is known to be y=(152−x2)12.

No, it is not. It would be if your circle was centered at the same origin you are using for the integration.

$\triangle POR= 12*9$
OPQR as a part of the circle =$15^2 *2 \arctan \frac{4}{3}$
Red area = $15^2 *2 \arctan \frac{4}{3} - 12*9$

OP already did this (”attempt 1”). He is asking for clarification where the calculus approach went wrong.

 

[brotherbobby]

No, it is not. It would be if your circle was centered at the same origin you are using for the integration.

Brilliant point. Thank you. Which means, were I to write the "equation of a circle" using a point other than the center as my origin, the equation would turn out to be nowhere as neat as $x^2+y^2=a^2$.
[I have found a way to do the problem using the second method. I'd be back soon with it].

 

[Delta2]

What you calculate as $A_{arc}$ is actually the rectangle with sides OQ and QS minus the green area PQS. Thus, If you subtract from $A_{arc}$ the rectangle with sides OT and TP you get the half of the red area=158-9x12=50 and thus the total red area is 2x50=100.

 

[brotherbobby]

I have solved the problem, partly my own realization and that of @Orodruin 's point above in post #3.

Problem Statement : To find the area of the shaded segment filled in red in the circle shown to the right. The region is marked by the points PQRP.

Attempt (using calculus) : I drew an "inverted" image of the problem in order facilitate its solution. I paste it to the left. Everything in white in the diagram is what's given. Everything in bright green is my own construction to help solve the problem.

The area of the segment RQTR in red is
$$A_{\text{seg}} = \text{Area of arc RQ with the horizontal - Area of line TQ with horizontal}$$
$A_{\text{seg}} = \int_0^{12}\sqrt{15^2-x^2}dx-12\times 9$
$A_{\text{seg}}\approx 158 - 108\; \text{see my calculations in post #1, second part}$
$A_{\text{seg}} \approx 50\text{ cm}^2$.

The entire segment PRQP in red is twice this segment; hence its area is $2\times 50 = \boxed{100}\;\text{cm}^2$.

 

[Orodruin]

Well, you have solved the problem. However, maybe not in the most convenient calculus manner.

Consider the following:
For any shape, the area between two $y$-values $a$ and $b$ is given by
$$
A(a,b) = \int_a^b \ell(y) dy
$$
where $\ell(y’)$ is the length of the line of constant $y=y’$ that lies within the shape. For a circle of radius $R$ centered at the origin
$$
\ell(y) = 2 \sqrt{R^2 - y^2}.
$$
Therefore, the area of the lowest cm of the circle with $R = 15$ cm is
$$
A(-15,-12) = 2 \int_{-15}^{-12} \sqrt{15^2 - y^2} dy.
$$
Computing this integral should give you the correct area.