Review question

# Can we sketch the graph of $y^2 = x^3$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6372

## Solution

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

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*}