The area contained between \(4 x=y^{2}\) and \(4 y=x^{2}\) is a) \(\frac{10}{3}\) b) \(\frac{11}{3}\) C) \(\frac{13}{3}\) d) \(\frac{16}{3}\)

To find the points of intersection between the curves, we need to solve the equation system: $$ \begin{cases} 4x = y^2 \\ 4y = x^2 \end{cases} $$ Let's solve it by substituting \(y^2\) from the first equation to the second equation: \((4x)^2 = 16x^2 = 4y\), or \(y = 4x^2\) Now substitute this expression of \(y\) in terms of \(x\) back into the first equation: \(4x = (4x^2)^2\), which simplifies to \(x^3 = 1\). Thus, \(x=1\) is the only solution, and the corresponding value for \(y\) is \(y = 4(1)^2 = 4\). Therefore, the two curves intersect at the point \((1,4)\).

Now that we have found the points of intersection, we'll use them as our limits of integration. To compute the area we'll use the vertical strip method of integration. Since the area is given in the first quadrant, we can write the integrals as: $$ \int_{0}^{1}(y_1 - y_2)dx $$ Where \(y_1\) and \(y_2\) represent the expressions of \(y\) in terms of \(x\) for the two given curves. We have: $$ y_1 = \sqrt{4x} \Rightarrow y_1 = 2\sqrt{x} $$ and $$ y_2 = 4x^2 $$ So the integral becomes: $$ \int_{0}^{1}(2\sqrt{x} - 4x^2)dx $$

Now, we will evaluate the integral: $$ \int_{0}^{1}(2\sqrt{x} - 4x^2)dx = \left[ \frac{4}{3} x^{\frac{3}{2}} - \frac{4}{3} x^3 \right]_0^1 $$ Evaluating the integral gives: $$ \left( \frac{4}{3} (1)^{\frac{3}{2}} - \frac{4}{3} (1)^3 \right) - \left( \frac{4}{3} (0)^{\frac{3}{2}} - \frac{4}{3} (0)^3 \right) = \frac{4}{3} - \frac{4}{3} = 0 $$ However, this result shows the net area is \(0\), which might be an indication that the integral limits are not well-defined or the integral is not set up correctly. Instead, we should define the integral in terms of \(y\): $$ \int_{0}^{4}(x_1 - x_2)dy $$ Now, we'll find the expressions of \(x\) in terms of \(y\) for the two given curves: $$ x_1 = \frac{y^2}{4} $$ and $$ x_2 = \sqrt{\frac{y}{4}} $$ The integral becomes: $$ \int_{0}^{4}\left(\frac{y^2}{4} - \sqrt{\frac{y}{4}}\right)dy $$

Let's evaluate the new integral: $$ \int_{0}^{4}\left(\frac{y^2}{4} - \sqrt{\frac{y}{4}}\right)dy = \left[\frac{y^3}{12} - \frac{2}{3}y^{\frac{3}{2}}\right]_0^4 $$ Evaluating the integral gives: $$ \left(\frac{4^3}{12} - \frac{2}{3}(4)^{\frac{3}{2}}\right) - \left(\frac{0^3}{12} - \frac{2}{3}(0)^{\frac{3}{2}}\right) = \frac{64}{3} - \frac{16}{3} = \frac{48}{3} = 16 $$ So the area contained between the two curves is \(16\), which corresponds to option (d).