Sketch the graph of \(y^2 = x^3\).

Since \(y^2 \ge 0\), the graph only exists for \(x^3 \ge 0\), that is, when \(x \ge 0\). Taking square roots we see that the curve has two symmetrical branches, namely \[\begin{equation*} y = x^{3/2} \quad\text{and}\quad y = -x^{3/2}. \end{equation*}\] It is clear that the positive branch increases. Its gradient is given by \[\begin{equation*} \frac{dy}{dx} = \frac{3}{2}x^{1/2} \end{equation*}\]

which is zero at \(x=0\) and increases with \(x\). Thus, we have

A sketch of the graph of y squared = x cubed.

Calculate the area enclosed by the curve and the ordinate \(x = 4\).

As the half-area above the \(x\)-axis is a reflection of the half-area below the \(x\)-axis, the area we want is \[\begin{equation*} 2 \int_0^4 x^{3/2} \:dx = 2 \left[ \frac{x^{5/2}}{\frac{5}{2}}\right]_0^4 = \frac{4}{5} \times 4^{5/2} = \frac{4}{5} \times 2^5 = \frac{128}{5}. \end{equation*}\]

Find also the volume generated by rotating this area about the axis of \(x\).

The volume of revolution obtained by rotating a curve around the \(x\)-axis is \[\begin{equation*} \int_{x_0}^{x_1} \pi y^2 \:dx. \end{equation*}\] So in our case, the volume is \[\begin{align*} \int_0^4 \pi x^3 \:dx = \pi \frac{4^4}{4} = 64\pi. \end{align*}\]