Review question

# Can we find this area by integrating with respect to $y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9815

## Solution

Sketch the form of the curve $27y^2=x^3$, …

The graph passes through the point $(0,0)$, and at no other point crosses the $x$- or $y$-axis.

The graph does not exist for negative $x$.

The curve will be symmetric about the $x$-axis, due to the $y^2$ term.

We can rearrange the equation for positive $y$ values as $y=\frac{1}{\sqrt{27}}x^{3/2}$.

Since $\dfrac{3}{2}>1$, this will be between the graph of $y=x$ and $y=x^2$, so it will look something like this…

… and find the equation of the tangent at the point $P(12,8)$ on the curve.

Differentiating implicitly, we find that $54y\frac{dy}{dx}=3x^2,$ and so $\frac{dy}{dx}=\frac{1}{18}\frac{x^2}{y}.$ Evaluating this at $P(12,8)$, the gradient of the tangent is $\frac{1}{18}\frac{12^2}{8}=\frac{144}{144}=1.$ So the tangent has the form $y=x+c.$

Since we know that the tangent passes through $(12,8)$, we find that $8=12+c \Longrightarrow c=-4.$ So the equation of the tangent to the curve at $P$ is $y=x-4.$

Verify this tangent cuts the curve again at the point $Q(3,-1)$.

For the line $y=x-4$, we have $-1=3-4=-1$ and so the line passes through $Q$.

For the curve, we have $27y^2=27\times (-1)^2=27\ \text{and}\ x^3=3^3=27.$

So the curve and the line intersect again at $Q(3,-1)$.

If $O$ is the origin, find by integration the area enclosed by the arcs $OP$, $OQ$ of the curve and the line $PQ$. [Integration with respect to $y$ is recommended.]

Here’s the area we want to find:

To find the shaded area, we find the area of the trapezium bounded by the lines $x=0, y=8$, $y=-1$ and $PQ$ and then subtract the area under the curve $x=3y^{2/3}$ between $y=-1$ and $y=8$.

The area of the trapezium is $\frac{12+3}{2}\times 9=\frac{135}{2}.$

The area under the curve is \begin{align*} \int_{-1}^8 3y^{2/3}\, dy &=\left[\frac{9}{5}y^{5/3}\right]^8_{-1}\\ &=\frac{9\times32}{5}+\frac{9}{5}\\ &=\frac{297}{5}. \end{align*}

So the area we want is $\frac{135}{2}-\frac{297}{5}=\frac{81}{10}.$