The area bounded by parabola \(y^{2}=x\), straight line \(y=4\) and \(y\)-axis is (1) \(\frac{16}{3}\) (2) \(\frac{64}{3}\) (3) \(7 \sqrt{2}\) (4) \(\frac{32}{3}\)

A parabola is a curve where any point is at an equal distance from a fixed point called the focus and a fixed straight line called the directrix. The equation \(y^2 = x\) describes a standard parabola that opens to the right. This means as you move further away from the origin in the positive direction along the x-axis, the values of y will either increase or decrease symmetrically. In this case, the vertex of the parabola is at the origin (0,0). Identifying the nature of the parabola and its orientation is crucial in solving problems related to bounded areas or intersections with other lines, as we'll see in the next sections.

The bounded area in this problem is defined by the parabola \(y^2 = x\), the line \(y = 4\), and the y-axis. To find this area, we need to determine the points where these boundaries intersect. The y-axis intersects the parabola at the origin \(0, 0\). The horizontal line \(y = 4\) intersects the parabola where \(y = 4\). By substituting \(y = 4\) into the parabolic equation, we get \(4^2 = x\), so \(x = 16\). Therefore, the point of intersection is (16, 4).
Once the bounds of the area are identified, we can represent the region of interest using definite integrals. It's the area between these bounds under the parabola, which forms the shape we are calculating.

A definite integral is an integral that calculates the net area under a curve between two specified limits. In this exercise, we need to integrate the function \(x = y^2 \) from \(y = 0\) to \(y = 4\) to find the area. The integral setup is given by:
\[ \int_{0}^{4} y^2 \, dy \]
Integrating \ y^2\ with respect to \ y\, we get:
\[ \int_{0}^{4} y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^4 = \frac{4^3}{3} - \frac{0^3}{3} = \frac{64}{3} \]
This computation concludes that the area bounded by the parabola, the line \(y = 4\), and the y-axis is \( \frac{64}{3} \text{square units}\). Definite integrals are a powerful tool for calculating the exact area under curves, making them essential for solving many mathematical and real-world problems.