## Task Condition

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

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

Find the ordinates of the points of intersection of the graphs of the functions

$x=\sqrt{4-y^{2}}, x=0$ :

$$
y= \pm 2
$$

In the case $x=\sqrt{4-y^{2}}, y=0$ it is obvious:

$$
y=0
$$

In the case $x=\sqrt{4-y^{2}}, y=1$ it is obvious:

$$
y=1
$$

Calculate the area:

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-53.jpg?height=988&width=1011&top_left_y=945&top_left_x=934)

$$
S=\int_{0}^{1}\left(\sqrt{4-y^{2}}-0\right) d y=\int_{0}^{1} \sqrt{4-y^{2}} d y=
$$

Substitution:

$$
\begin{aligned}
& y=2 \sin t \Rightarrow d y=2 \cos t d t \\
& t=\arcsin \frac{y}{2} \\
& y=0 \Rightarrow t=\arcsin \frac{0}{2}=0 \\
& y=1 \Rightarrow t=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \sqrt{4-4 \sin ^{2} t} \cdot 2 \cos t d t=4 \cdot \int_{0}^{\pi / 6} \sqrt{1-\sin ^{2} t} \cdot \cos t d t=4 \cdot \int_{0}^{\pi / 6} \sqrt{\cos ^{2} t} \cdot \cos t d t= \\
& =4 \cdot \int_{0}^{\pi / 6} \cos ^{2} t d t=4 \cdot \int_{0}^{\pi / 6} \frac{1}{2} \cdot(\cos 2 t+1) d t=2 \cdot \int_{0}^{\pi / 6}(\cos 2 t+1) d t= \\
& =\left.2 \cdot\left(\frac{1}{2} \sin 2 t+t\right)\right|_{0} ^{\pi / 6}=\left.(\sin 2 t+2 t)\right|_{0} ^{\pi / 6}=\left(\sin \frac{\pi}{3}+\frac{\pi}{3}\right)-(\sin 0+0)=\frac{\sqrt{3}}{2}+\frac{\pi}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-28$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

- Last edited: 07:38, June 9, 2010.
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## Problem Kuznetsov Integrals 14-29

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