## Problem Statement

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

$$
y=x \sqrt{36-x^{2}}, y=0, (0 \leq x \leq 6)
$$
## Solution

Integrals 14-13

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-24.jpg?height=1425&width=1430&top_left_y=1041&top_left_x=450)

$S=\int_{0}^{6} x \sqrt{36-x^{2}} d x=$

$$
\begin{aligned}
& =\frac{1}{2} \int_{0}^{6} \sqrt{36-x^{2}} d\left(x^{2}\right)= \\
& =-\frac{1}{2} \int_{0}^{6}\left(36-x^{2}\right)^{1 / 2} d\left(36-x^{2}\right)= \\
& =-\left.\frac{1}{2}\left(\frac{2}{3} \cdot\left(36-x^{2}\right)^{3 / 2}\right)\right|_{0} ^{6}= \\
& =-\frac{1}{3}\left((36-36)^{3 / 2}-(36-0)^{3 / 2}\right)=72
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-13$ " Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-14

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