Can we sketch the graph of $y^2 = x^3$?

Can we sketch the graph of $y^2 = x^3$?

Question

Sketch the graph of $y^2 = x^3$.

Calculate the area enclosed by the curve and the ordinate $x = 4$.

Find also the volume generated by rotating this area about the axis of $x$.

Solution

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

As the half-area above the

$x$

-axis is a reflection of the half-area below the

$x$

-axis, the area we want is

$$\begin{equation*} 2 \int_0^4 x^{3/2} \:dx = 2 \left[ \frac{x^{5/2}}{\frac{5}{2}}\right]_0^4 = \frac{4}{5} \times 4^{5/2} = \frac{4}{5} \times 2^5 = \frac{128}{5}. \end{equation*}$$

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

$$\begin{align*} \int_0^4 \pi x^3 \:dx = \pi \frac{4^4}{4} = 64\pi. \end{align*}$$