## Problem Statement

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

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

Let's find the points of intersection:

$$
\begin{aligned}
& \text { Integrals 15-28 } \\
& \left\{\begin{array}{l}
x=4 \sqrt{2} \cdot \cos ^{3} t \\
y=\sqrt{2} \cdot \sin ^{3} t
\end{array}\right.
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-59.jpg?height=325&width=1422&top_left_y=1311&top_left_x=497)

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

Since the functions \( x=4 \sqrt{2} \cdot \cos ^{3} t \) and \( y=\sqrt{2} \cdot \sin ^{3} t \) are periodic (with a period of \( 2 \pi \)), we can take any interval of length \( 2 \pi \). Let's take \([- \pi ; \pi]\). Then:

$$
\begin{aligned}
& t=-\frac{\pi}{4} \text { or } t=\frac{\pi}{4} \\
& x \geq 2 \text { on the interval }\left[-\frac{\pi}{4} ; \frac{\pi}{4}\right] .
\end{aligned}
$$

From the graph, it is clear that the region is symmetric with respect to the \( x \)-axis, and its area can be calculated using the formula:

$$
\begin{aligned}
& S=-2 \cdot \int_{0}^{\pi / 4} y(t) \cdot x^{\prime}(t) d t= \\
& =-2 \cdot \int_{0}^{\pi / 4} \sqrt{2} \sin ^{3} t \cdot\left(4 \sqrt{2} \cos ^{3} t\right)^{\prime} d t= \\
& =-2 \cdot \int_{0}^{\pi / 4} \sqrt{2} \sin ^{3} t \cdot 12 \sqrt{2} \cos ^{2} t \cdot(-\sin t) d t= \\
& =48 \cdot \int_{0}^{\pi / 4} \sin ^{4} t \cdot \cos ^{2} t d t=\left|\begin{array}{l}
\sin t \cos t=\frac{1}{2} \sin 2 t \\
\sin ^{2} t=\frac{1}{2}(1-\cos 2 t)
\end{array}\right|= \\
& =48 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 2 t) \cdot\left(\frac{1}{2} \sin 2 t\right)^{2} d t=6 \cdot \int_{0}^{\pi / 4}(1-\cos 2 t) \sin ^{2} 2 t d t= \\
& =6 \cdot \int_{0}^{\pi / 4} \sin ^{2} 2 t d t-6 \cdot \int_{0}^{\pi / 4} \cos 2 t \cdot \sin ^{2} 2 t d t= \\
& =6 \cdot \int_{0}^{\pi / 4} \frac{1}{2}(1-\cos 4 t) d t-6 \cdot \int_{0}^{\pi / 4} \frac{1}{2} \cdot \sin ^{2} 2 t d(\sin 2 t)= \\
& =\left.3 \cdot\left(t-\frac{1}{4} \sin 4 t\right)\right|_{0}^{\pi / 4}-\left.3 \cdot\left(\frac{1}{3} \sin ^{3} 2 t\right)\right|_{0}^{\pi / 4}= \\
& =3 \cdot\left(\frac{\pi}{4}-\frac{1}{4} \sin \frac{4 \pi}{4}\right)-3 \cdot\left(0-\frac{1}{4} \sin 0\right)-\left(\sin ^{3} \frac{2 \pi}{4}-\sin ^{3} 0\right)= \\
& =\frac{3 \pi}{4}-\frac{3}{4} \cdot \sin \pi-0-\sin ^{3} 2 \pi+0= \\
& =\frac{3 \pi}{4}-0-0^{3}=\frac{3 \pi}{4}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-28$ » Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals | Problems for Checking

- Last edited: 11:05, June 17, 2010.
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## Problem Kuznetsov Integrals 15-29

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