334. Calculate the area of the figure bounded by the curves $y^{2}=4 x$ and $x^{2}=4 y$
Solution. In Fig. 182, the area of the figure we need to calculate is shown. As can be seen from the figure, the area of the figure OBM AO can be represented as the difference between the areas of the figures OBMPO and OAMPO, where $MP$ is the perpendicular dropped from point $M$ to the $Ox$ axis.

Let's find the coordinates of point $M$. By solving the system of equations $\left\{\begin{array}{l}y^{2}=4 x \\ y=x^{2} / 4\end{array}\right.$, we get $x=4, y=4$, i.e., $M(4 ; 4)$.

Therefore,

$$
S=\int_{0}^{4} \sqrt{4} x d x-\int_{0}^{4} \frac{x^{2}}{4} d x=\left.\frac{4}{3} x^{3 / 2}\right|_{0} ^{4}-\left.\frac{x^{3}}{12}\right|_{0} ^{4}=\frac{16}{3} \text { (sq. units) }
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-330.jpg?height=269&width=402&top_left_y=1533&top_left_x=214)

Fig. 183

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-330.jpg?height=319&width=346&top_left_y=1481&top_left_x=704)

Fig. 184

![](https://cdn.mathpix.com/cropped/2024_05_22_db4d450c77c65a914ec1g-331.jpg?height=287&width=371&top_left_y=187&top_left_x=107)

Fig. 185

This problem can also be solved in another way. Represent the desired area as the difference between the areas of the figures OAMNO and $O B M N O \quad(M N$ - the perpendicular dropped from point $M$ to the $Oy$ axis), i.e., $S=S_{\text {OAMNO- }}$ $-S_{\text {Oвмnо: }}$ Then

$$
\begin{gathered}
S=\int_{0}^{4} \sqrt{4 y} d y-\int_{0}^{4} \frac{y^{2}}{4} d y=\left.\frac{4}{3} y^{3 / 2}\right|_{0} ^{4}-\left.\frac{y^{3}}{12}\right|_{0} ^{4}= \\
=\frac{16}{3} \text { (sq. units). }
\end{gathered}
$$