- #1
brotherbobby
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- 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}}##.
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.