## Task Condition

Calculate the area of the figure bounded by the graphs of the functions:

$$
y=\sqrt{4-x^{2}}, y=0, x=0, x=1
$$
## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-08.jpg?height=1133&width=1514&top_left_y=1141&top_left_x=408)

The desired area \( S \) is:

\( S=\int_{0}^{1} \sqrt{4-x^{2}} d x \)

We make a trigonometric substitution:
\( \sqrt{4-x^{2}}=2 \sin t \), hence \( x=2 \cos t, d x=2 \sin t \)

When \( x=0=2 \cos t \Rightarrow t=\frac{\pi}{2} \)

and when \( x=1=2 \cos t \Rightarrow t=\frac{\pi}{3} \)

Then we get

\[
\begin{aligned}
& S=\int_{0}^{1} \sqrt{4-x^{2}} d x= \\
& =\int_{\pi / 3}^{\pi / 2} 2 \sin t \cdot 2 \sin t d t= \\
& =4 \int_{\pi / 3}^{\pi / 2} \sin ^{2} t d t= \\
& =2 \int_{\pi / 3}^{\pi / 2}(1-\cos 2 t) d t= \\
& =\left.(2 t-\sin 2 t)\right|_{\pi / 3} ^{\pi / 2}=\frac{\pi}{3}+\frac{\sqrt{3}}{2}
\end{aligned}
\]

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-5$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-6

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