## Problem Statement

Calculate the area of the figure bounded by the lines given by the equations.

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8 \sqrt{2} \cdot \cos ^{3} t \\
y=\sqrt{2} \cdot \sin ^{3} t
\end{array}\right. \\
& x=4(x \geq 4)
\end{aligned}
$$
## Solution

Let's find the points of intersection:

$$
\begin{aligned}
& x=8 \sqrt{2} \cdot \cos ^{3} t=4 \\
& \cos ^{3} t=\frac{1}{2 \sqrt{2}} \\
& \cos t=\frac{1}{\sqrt{2}} \\
& t= \pm \frac{\pi}{4}+2 \pi k, k \in \mathbb{Z}
\end{aligned}
$$

Since the functions

$x=8 \sqrt{2} \cdot \cos ^{3} t, y=\sqrt{2} \cdot \sin ^{3} t$

are periodic (with period $2 \pi$), we can take any interval of length $2 \pi$. Let's take $[-\pi ; \pi]$. Then:

$$
\begin{aligned}
& t=-\frac{\pi}{4} \text { or } \\
& t=\frac{\pi}{4} \\
& x \geq 4 \text { on the interval }\left[-\frac{\pi}{4} ; \frac{\pi}{4}\right] .
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-19.jpg?height=1017&width=1005&top_left_y=1096&top_left_x=925)

We calculate the area:

$$
\begin{aligned}
& S=\int_{-\pi / 4}^{\pi / 4} 8 \sqrt{2} \cdot \cos ^{3} t \cdot\left(\sqrt{2} \cdot \sin ^{3} t\right)^{\prime} d t=16 \cdot \int_{-\pi / 4}^{\pi / 4} \cos ^{3} t \cdot 3 \sin ^{2} t \cdot \cos t d t= \\
& =48 \cdot \int_{-\pi / 4}^{\pi / 4} \cos ^{4} t \cdot \sin ^{2} t d t=\left|\sin t \cos t=\frac{1}{2} \sin 2 t, \cos ^{2} t=\frac{1}{2}(1+\cos 2 t)\right|=
\end{aligned}
$$

$$
\begin{aligned}
& =48 \cdot \int_{-\pi / 4}^{\pi / 4} \frac{1}{2}(1+\cos 2 t) \cdot\left(\frac{1}{2} \sin 2 t\right)^{2} d t=6 \cdot \int_{-\pi / 4}^{\pi / 4}(1+\cos 2 t) \sin ^{2} 2 t d t= \\
& =6 \cdot \int_{-\pi / 4}^{\pi / 4} \sin ^{2} 2 t d t+6 \cdot \int_{-\pi / 4}^{\pi / 4} \cos 2 t \cdot \sin ^{2} 2 t d t=\left|\sin ^{2} 2 t=\frac{1}{2}(1-\cos 4 t)\right|= \\
& =6 \cdot \int_{-\pi / 4}^{\pi / 4} \frac{1}{2}(1-\cos 4 t) d t+6 \cdot \int_{-\pi / 4}^{\pi / 4} \frac{1}{2} \cdot \sin ^{2} 2 t d(\sin 2 t)= \\
& =\left.3 \cdot\left(t-\frac{1}{4} \sin 4 t\right)\right|_{-\pi / 4} ^{\pi / 4}+\left.3 \cdot\left(\frac{1}{3} \sin ^{3} 2 t\right)\right|_{-\pi / 4} ^{\pi / 4}= \\
& =3 \cdot\left(\frac{\pi}{4}-\frac{1}{4} \sin \pi\right)-3 \cdot\left(-\frac{\pi}{4}-\frac{1}{4} \sin (-\pi)\right)+\sin ^{3} \frac{\pi}{2}-\sin ^{3}\left(-\frac{\pi}{2}\right)= \\
& =3 \cdot\left(\frac{\pi}{4}-\frac{1}{4} \cdot 0\right)-3 \cdot\left(-\frac{\pi}{4}-\frac{1}{4} \cdot 0\right)+1^{3}-(-1)^{3}= \\
& =\frac{3 \pi}{4}-0+\frac{3 \pi}{4}+0+1+1=\frac{3 \pi}{2}+2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-10$ » Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals

- Last edited: 07:39, 16 June 2010.
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## Problem Kuznetsov Integrals 15-11

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