Area of segment cut from a circle

[brotherbobby]

Homework Statement

Statement : From a circle of radius $a$, a segment is cut off at a height $h$ from the base, as shown in the figure below. Calculate the area of the region shaded in $\color{red}{red}$.

Relevant Equations

1. The area of the sector of a circle that subtends an angle of $\theta^c$ at the center is $A_{\odot} = \dfrac{1}{2}a^2\theta^c$.



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



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



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



5. The integral $\displaystyle{\int\sqrt{a^2-x^2}\;{\rm{d}}x = \dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}}$.

I had posed this problem as the OP here . I am trying to do the same problem here, without numbers.

Drawing : I put the image of the problem to the right. O is the center of the circle distant $a$ from the circumference like points P and R. At a height $h$ from the base Q, a segment is cut and shaded in red. I have to calculate the area of the shaded region RPQR.

Attempt : I do the problem in two ways, one using the methods of algebra and trigonometry and the next using calculus. The trouble is, my answers don't match.

${\color{blue}{\mathbf{I. Algebra}} :}$ From the diagram alongside, $\cos\theta = \frac{a-h}{a}=1-\frac{h}{a}$. Area of sector ORQPO : $\mathcal{A}_{\text{sector}}(\text{ORQPO}) = \dfrac{1}{2}a^2(2\theta) = a^2\theta=a^2\cos^{-1}\left(1-\frac{h}{a}\right)\qquad{\color{blue}{\mathbf{(1)}}}$

The base of $\triangle OPT = b = \sqrt{a^2-(a-h)^2}=\sqrt{2ah-h^2}$.

Area of triangle ORPO : $\mathcal{A}_{\triangle} (\text{ORPO}) = \frac{1}{2}\times 2b\times (a-h)=b(a-h)=(a-h)\sqrt{2ah-h^2}\qquad{\color{blue}{\mathbf{(2)}}}$

Using (1) and (2), area of the segment RTPQR : $\boxed{\mathcal{A}_{\text{segment}}(\text{RTPQR})} = \mathcal{A}_{\text{sector}} - \mathcal{A}_{\triangle} = \boxed{a^2\cos^{-1}\left(1-\frac{h}{a}\right) - (a-h)\sqrt{2ah-h^2}}\qquad{\color{blue}{\mathbf{(3)}}}$

${\color{blue}{\mathbf{II. Calculus}} :}$ I "flip" the image horizontally and draw it to the right. The right side of the shaded region is included by the arc in green QP and the horizontal line TP.

Area of the region RQPMNR is : $\small{\mathcal{A}(\text{RQPMNR}) = \displaystyle{2\int\limits_0^b \sqrt{a^2-x^2}dx = \left.x\sqrt{a^2-x^2}\right|_0^b+\left. a^2\sin^{-1}\dfrac{x}{a}\right|_0^b} = b\sqrt{a^2-b^2}+a^2\sin^{-1}\dfrac{b}{a} = \sqrt{2ah-h^2}\sqrt{a^2-2ah+h^2}+a^2\sin^{-1}\dfrac{\sqrt{2ah-h^2}}{a}}$, upon putting the value of $b$ found above in $\mathbf{I}$.
This simplifies to give the area $\small{\mathcal{A}(\text{RQPMNR}) = \sqrt{2ah-h^2}(a-h)+a^2\sin^{-1}\left(\dfrac{h}{a}\sqrt{\dfrac{2a}{h}-1}\right)}\qquad$ (4)

Area of the rectangle RPMNR is : $\small{\mathcal{A}_{\sqsubset\!\sqsupset} = 2(a-h)b=2(a-h)\sqrt{2ah-h^2}}\qquad$ (5)
Using (4) and (5), area of the segment RTPQR is :

$\boxed{\mathcal{A}_{\text{segment}}(\text{RTPQR})=a^2\sin^{-1}\left(\dfrac{h}{a}\sqrt{\dfrac{2a}{h}-1}\right)-(a-h)\sqrt{2ah-h^2}}\quad$ (6)

The expressions in (3) and (6) would be the same if $\quad\underline{\cos^{-1}\left(1-\dfrac{h}{a}\right) = \sin^{-1}\left(\dfrac{h}{a}\sqrt{\dfrac{2a}{h}-1}\right)}$.

The only thing I know from inverse trigonometric functions is that $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$. Clearly, this doesn't reduce from expression to the other in the underlined equation above.

Request : A hint regarding the underlined equation above involving the inverse cosine and sine expressions.

 

[pbuk]

I think it would be easier to find expressions in terms of $\theta$.

And why on Earth would you use the letter $a$ to denote the radius of a circle?

 

[brotherbobby]

And why on Earth would you use the letter a to denote the radius of a circle?

I think it's a small point. Have you gone through Gregory's book on Classical Mechanics (2006)? He has used the letter $b$ for the radius of circles, keeping the letter $a$ for quantities like acceleration.

All the same, I have done the problem, that is, I have shown that $\quad\underline{\cos^{-1}\left(1-\dfrac{h}{a}\right) = \sin^{-1}\left(\dfrac{h}{a}\sqrt{\dfrac{2a}{h}-1}\right)}$.

I did this by taking a right angled triangle and using the Pythagorean theorem, whereby the angle $\theta=\cos^{-1}\left(1-\dfrac{h}{a}\right)$ turns out the same as the angle given by $\sin^{-1}\left(\dfrac{h}{a}\sqrt{\dfrac{2a}{h}-1}\right)$.

Silly from my side, again. Sorry to waste your time, forget about mine.