## Problem Statement

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
x=8 \cos ^{3} t \\
y=8 \sin ^{3} t
\end{array}\right. \\
& x=1(x \geq 1)
\end{aligned}
$$
## Solution

Let's find the points of intersection:

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

Since the functions

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

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

$$
\begin{aligned}
& t=-\frac{\pi}{3} \text { or } t=\frac{\pi}{3} \\
& x \geq 1 \text { on the interval }\left[-\frac{\pi}{3} ; \frac{\pi}{3}\right] .
\end{aligned}
$$

From the graph, it is clear that the region is symmetric with respect to the $0 x$ axis, and its area can be calculated using the formula:

$$
\begin{aligned}
& S=-2 \cdot \int_{0}^{\pi / 3} y(t) \cdot x^{\prime}(t) d t= \\
& =-2 \cdot \int_{0}^{\pi / 3} 8 \sin ^{3} t \cdot\left(8 \cos ^{3} t\right)^{\prime} d t=
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa23e2deef6f8d90c6d5g-46.jpg?height=1344&width=973&top_left_y=1087&top_left_x=947)

$$
\begin{aligned}
& =-2 \cdot \int_{0}^{\pi / 3} 8 \sin ^{3} t \cdot 24 \cos ^{2} t \cdot(-\sin t) d t= \\
& =384 \cdot \int_{0}^{\pi / 3} \sin ^{4} t \cdot \cos ^{2} t d t=\left|\begin{array}{l}
\sin t \cos t=\frac{1}{2} \sin 2 t \\
\sin ^{2} t=\frac{1}{2}(1-\cos 2 t)
\end{array}\right|= \\
& =384 \cdot \int_{0}^{\pi / 3} \frac{1}{2}(1-\cos 2 t) \cdot\left(\frac{1}{2} \sin 2 t\right)^{2} d t=48 \cdot \int_{0}^{\pi / 3}(1-\cos 2 t) \sin ^{2} 2 t d t= \\
& =48 \cdot \int_{0}^{\pi / 3} \sin ^{2} 2 t d t-48 \cdot \int_{0}^{\pi / 3} \cos 2 t \cdot \sin ^{2} 2 t d t= \\
& =48 \cdot \int_{0}^{\pi / 3} \frac{1}{2}(1-\cos 4 t) d t-48 \cdot \int_{0}^{\pi / 3} \frac{1}{2} \cdot \sin ^{2} 2 t d(\sin 2 t)= \\
& =\left.24 \cdot\left(t-\frac{1}{4} \sin 4 t\right)\right|_{0} ^{\pi / 3}-\left.24 \cdot\left(\frac{1}{3} \sin ^{3} 2 t\right)\right|_{0} ^{\pi / 3}= \\
& =24 \cdot\left(\frac{\pi}{3}-\frac{1}{4} \sin \frac{4 \pi}{3}\right)-24 \cdot\left(0-\frac{1}{4} \sin ^{2}\right)-8 \cdot\left(\sin ^{3} \frac{2 \pi}{6}-\sin ^{3} 0\right)= \\
& =8 \pi-6 \cdot \sin \left(\pi+\frac{\pi}{3}\right)-0-8 \cdot \sin ^{3} \frac{\pi}{3}+0= \\
& =8 \pi-6 \cdot\left(-\sin \frac{\pi}{3}\right)-8 \cdot \sin ^{3} \frac{\pi}{3}= \\
& =8 \pi+6 \cdot \frac{\sqrt{3}}{2}-8 \cdot\left(\frac{\sqrt{3}}{2}\right)^{3}=8 \pi+\frac{\sqrt{3}}{2} \cdot\left(6-8 \cdot \frac{3}{4}\right)=8 \pi+\frac{\sqrt{3}}{2} \cdot 0=8 \pi
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+15-22$ »

Categories: Kuznetsov's Problem Book Integrals Problem 15 | Integrals | Problems for Checking

- Last edited: 12:22, June 17, 2010.
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## Problem Kuznetsov Integrals $15-23$

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