## Problem Statement

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

$$
y=x^{2} \sqrt{8-x^{2}}, y=0, (0 \leq x \leq 2 \sqrt{2})
$$
## Solution

The desired area $S$ is:

$S=\int_{0}^{2 \sqrt{2}} x^{2} \sqrt{8-x^{2}} d x$

We will use a trigonometric substitution:

$\sqrt{8-x^{2}}=2 \sqrt{2} \sin t$, hence

$x=2 \sqrt{2} \cos t, d x=2 \sqrt{2} \sin t d t$

When $x=0=2 \sqrt{2} \cos t \Rightarrow t=\frac{\pi}{2}$

and when $x=1=2 \sqrt{2} \cos t \Rightarrow t=0$

Then we get

$S=\int_{0}^{2 \sqrt{2}} x^{2} \sqrt{8-x^{2}} d x$

Integrals 14-16

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-29.jpg?height=1422&width=788&top_left_y=1042&top_left_x=1139)

$=\int_{0}^{\pi / 2}(2 \sqrt{2} \cos t)^{2} \cdot \sqrt{8-(2 \sqrt{2} \cos t)^{2}} \cdot 2 \sqrt{2} \sin t d t=$

$$
\begin{aligned}
& =\int_{0}^{\pi / 2} 4 \cdot 2 \cos ^{2} t \cdot \sqrt{8-4 \cdot 2 \cos ^{2} t} \cdot 2 \sqrt{2} \sin t d t= \\
& =16 \sqrt{2} \int_{0}^{\pi / 2} \cos ^{2} t \cdot 2 \sqrt{2} \cdot \sqrt{1-\cos ^{2} t} \cdot \sin t d t= \\
& =64 \int_{0}^{\pi / 2} \cos ^{2} t \cdot \sin ^{2} t d t= \\
& =\frac{64}{4} \int_{0}^{\pi / 2} \sin ^{2} 2 t d t= \\
& =\frac{16}{2} \int_{0}^{\pi / 2}(1-\cos 4 t) d t= \\
& =\left.(8 t-2 \sin 4 t)\right|_{0} ^{\pi / 2}=8 \cdot \frac{\pi}{2}-2 \cdot \sin \frac{4 \pi}{2}=4 \pi-0=4 \pi
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_14-16" Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-17

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