## Problem Statement

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

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

Integral $14-6$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-10.jpg?height=1425&width=1430&top_left_y=1044&top_left_x=447)

$S=\int_{0}^{2} x^{2} \sqrt{4-x^{2}} d x=\left|\begin{array}{c}x=2 \sin t \\ d x=2 \cos t d t \\ x=0 ; t=0 \\ x=2 ; t=\pi / 2\end{array}\right|=$

$$
\begin{aligned}
& =\int_{0}^{\pi / 2} 4 \sin ^{2} t \sqrt{4-4 \sin ^{2} t} \cdot 2 \cos t d t= \\
& =4 \cdot 2 \cdot 2 \int_{0}^{\pi / 2} \sin ^{2} t \cdot \cos t \cdot \cos t d t= \\
& =16 \int_{0}^{\pi / 2} \sin ^{2} t \cdot \cos ^{2} t d t= \\
& =16 \int_{0}^{\pi / 2}\left(1-\cos ^{2} t\right) \cdot \cos ^{2} t d t= \\
& =16 \int_{0}^{\pi / 2} \cos ^{2} t d t-16 \int_{0}^{\pi / 2} \cos ^{4} t d t= \\
& =16 \int_{0}^{\pi / 2} \frac{1+\cos 2 t}{2} d t-16 \int_{0}^{\pi / 2} \frac{1+2 \cos 2 t+\cos ^{2} 2 t}{4} d t= \\
& =8 \int_{0}^{\pi / 2}(1+\cos 2 t) d t-4 \int_{0}^{\pi / 2}\left(1+2 \cos 2 t+\cos ^{2} 2 t\right) d t= \\
& =8 \int_{0}^{\pi / 2} d t+8 \int_{0}^{\pi / 2} \cos 2 t d t-4 \int_{0}^{\pi / 2} d t-4 \int_{0}^{\pi / 2} 2 \cos 2 t d t-4 \int_{0}^{\pi / 2} \cos ^{2} 2 t d t= \\
& =4 \int_{0}^{\pi / 2} d t+8 \int_{0}^{\pi / 2} \cos 2 t d t-8 \int_{0}^{\pi / 2} \cos 2 t d t-4 \int_{0}^{\pi / 2} \frac{1+\cos 4 t}{2} d t= \\
& =4 \int_{0}^{\pi / 2} d t-2 \int_{0}^{\pi / 2} d t-2 \int_{0}^{\pi / 2} \cos 4 t d t= \\
& =2 \int_{0}^{\pi / 2} d t-\frac{2}{4} \int_{0}^{\pi / 2} \cos 4 t d(4 t)= \\
& =\left.2 t\right|_{0} ^{\pi / 2}-\left.\frac{1}{2} \sin 4 t\right|_{0} ^{\pi / 2}= \\
& =2 \cdot \frac{\pi}{2}-0-\frac{1}{2}(\sin 2 \pi-\sin 0)=\pi-\frac{1}{2}(0-0)=\pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-6 \%$

Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

- Last edited: 06:43, June 10, 2010.
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## Problem Kuznetsov Integrals $14-7$

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