1.8. Calculate the areas of figures bounded by lines in the Cartesian coordinate system:

a) $3 x^{2}=25 y ; 5 y^{2}=9 x$

b) $x y=4 ; x+y=5$

c) $y=e^{x} ; y=e^{2 x} ; x=1$

d) $x+y=1 ; x+3 y=1 ; x=y ; x+2 y=2$
Solution. To calculate the areas of the specified figures $\Omega$, we will use formula (1.76)

$$
\mu(\Omega)=S_{\Omega 2}=\iint_{\Omega} d x d y
$$

a) First, we solve the equations $3 x^{2}=25 y$, $5 y^{2}=9 x$ simultaneously. Multiplying them, we get

$$
(x y)^{2}=15 x y .
$$

Thus, either $x y=0$ or $x y=15$. In the first case, we get the point $O(0,0)$, and in the second case, the point $x=5, y=3$.

We calculate the area using repeated integration:

$$
\begin{gathered}
S_{\Omega}=\iint_{\Omega} d x d y=\int_{0}^{5} d x \int_{\frac{3}{25} x}^{3 \sqrt{\frac{x}{5}}} d y= \\
=\int_{0}^{5}\left(\frac{3}{\sqrt{5}} x^{\frac{1}{2}}-\frac{3}{25} x^{2}\right) d x=\left.\frac{2}{\sqrt{5}} x^{\frac{3}{2}}\right|_{0} ^{5}-\left.\frac{x^{3}}{25}\right|_{0} ^{5}=7
\end{gathered}
$$

b) In this case, we solve the equations $x y=4$, $x+y=5$ simultaneously to determine the projections of the figure $\Omega$ on the coordinate axes. This gives us two points: $x=1 ; y=4$ and $x=4$, $y=1$. Now we calculate the area using repeated integration:

$$
\begin{aligned}
S_{\Omega}=\iint_{\Omega} d x d y & =\int_{1}^{4} d y \int_{\frac{4}{y}}^{5-y} d x=\int_{1}^{4}\left(5-y-\frac{4}{y}\right) d y=\left.\left(5-\frac{y^{2}}{2}-4 \ln y\right)\right|_{1} ^{4}= \\
& =20-8-8 \ln 2-5+\frac{1}{2}=\frac{15}{2}-8 \ln 2
\end{aligned}
$$

c) In this case, the figure $\Omega$ projects onto the $O X$ axis as the interval $[0,1]$, so

$$
\begin{gathered}
S_{\Omega}=\iint_{\Omega} d x d y=\int_{0}^{1} d x \int_{e^{x}}^{e^{2 x}} d y=\int_{0}^{1}\left(e^{2 x}-e^{x}\right) d x=\left.\left(\frac{1}{2} e^{2 x}-e^{x}\right)\right|_{0} ^{1}= \\
=\frac{e^{2}}{2}-e-\frac{1}{2}+1=\frac{e^{2}-2 e+1}{2}=\frac{(e-1)^{2}}{2}
\end{gathered}
$$

d) Let's draw the figure $\Omega$ whose area we need to determine (see Fig. 1.33).

Through points $C$ and $D$, draw lines parallel to the $O X$ axis. The ordinate of point $C: y=\frac{1}{4}$, the ordinate of point $A: y=\frac{1}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-053.jpg?height=328&width=561&top_left_y=129&top_left_x=393)

Fig. 1.33

the ordinate of point $D: y=\frac{2}{3}$. The region $\Omega$ consists of two triangles $A B C$ and $A D P$. We project them onto the $O Y$ axis and apply repeated integration:

$$
\begin{aligned}
& S_{\Omega}=\iint_{\Omega} d x d y=\int_{0}^{\frac{1}{4}} d y \int_{1-3 y}^{1} d x+\int_{\frac{1}{4}}^{\frac{1}{2}} d y \int_{y}^{1-y} d x+\int_{\frac{1}{2}}^{\frac{2}{3}} d y \int_{1-y}^{y} d x+\int_{\frac{2}{3}}^{1} d y \int_{1-y}^{2-2} d x= \\
& =\int_{0}^{\frac{1}{4}} 2 y d y+\int_{\frac{1}{4}}^{\frac{1}{2}}(1-2 y) d y+\int_{\frac{1}{2}}^{\frac{2}{3}}(2 y-1) d y+\int_{\frac{2}{3}}^{1}(1-y) d y= \\
& =\left.y^{2}\right|_{0} ^{\frac{1}{4}}+\left.\left(y-y^{2}\right)\right|_{\frac{1}{4}} ^{\frac{1}{2}}+\left.\left(y^{2}-y\right)\right|_{\frac{1}{2}} ^{\frac{2}{3}}+\left.\left(y-\frac{y^{2}}{2}\right)\right|_{\frac{2}{3}} ^{1}= \\
& =\frac{1}{16}+\frac{1}{4}-\frac{1}{16}+\frac{4}{9}-\frac{2}{3}-\frac{1}{4}+\frac{1}{2}+1-\frac{1}{2}-\frac{2}{3}+\frac{2}{9}=\frac{11}{12} .
\end{aligned}
$$