The area of a circle by slicing
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Solution
The main idea is to divide a circle into rectangular slices.
\[\begin{gathered}
R^2=x_i^2+y_i^2 \\\\
x_i=\sqrt{R^2-y_i^2}
\end{gathered}\]
The area of the \(i^{\text {th}} \) slice is
\[A_i=2 x_i \Delta y=2 \sqrt{R^2-y_i^2} \Delta y\]
Now let’s calculate the area of the circle:
\[A=\lim _{n \rightarrow+\infty} \sum_{i=1}^n A_i=\lim _{n \rightarrow+\infty} \sum_{i=1}^n 2 \sqrt{R^2-y_i^2} \Delta y=\int_{-R}^R 2 \sqrt{R^2-y^2} d y\]
Let
\[\begin{aligned}
& y=R \sin \theta \\\\
& d y=R \cos \theta d \theta
\end{aligned}\]
\[\begin{aligned}
& \mathrm{A}=2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{R^2-R^2 \sin ^2 \theta} \mathrm{R} \cos \theta d \theta=2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{R^2\left(1-\sin ^2 \theta\right)} \mathrm{R} \cos \theta d \theta \\\\
& =2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} R \sqrt{\cos ^2 \theta} \mathrm{R} \cos \theta d \theta=2 R^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos ^2 \theta d \theta \\\\
& \cos ^2 \theta=\frac{1}{2}(1+\cos 2 \theta) \\\\
& A=2 R^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(1+\cos 2 \theta) d \theta \\\\
& =R^2\left[\theta+\frac{\sin 2 \theta}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\\\
& =\pi R^2 \\\\
& A=\pi R^2
\end{aligned}\]
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