Review question

# Can we find the area bounded by these two graphs? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9386

## Solution

The area bounded by the graphs $y=\sqrt{2-x^2} \qquad \text{and} \qquad x+\left(\sqrt{2}-1\right)y=\sqrt{2}$ equals

1. $\dfrac{\sin{\sqrt{2}}}{\sqrt{2}}$;
2. $\dfrac{\pi}{4}-\dfrac{1}{\sqrt{2}}$;
3. $\dfrac{\pi}{2\sqrt{2}}$;
4. $\dfrac{\pi^2}{6}$.

The equation $y=\sqrt{2-x^2}$ can be squared to get $y^2=2-x^2$, or $x^2+y^2=2$. This is a circle centred at the origin with radius $\sqrt{2}$.

The original equation, though, is $y=\sqrt{\vphantom{x^2}\cdots}$, which is never negative, so we actually only have the semicircle with $y\ge0$.

The graph of $x+\left(\sqrt{2}-1\right)y=\sqrt{2}$ is a straight line which intersects the $y$-axis at $\left(0, 2+\sqrt{2}\right)$, and the $x$-axis at $\left(\sqrt{2},0\right)$, which is on the circumference of the circle.

We can now sketch these two graphs.

To find the other point of intersection, we substitute $x=\sqrt{2}-\left(\sqrt{2}-1\right)y$ into the equation of the circle, to get $\begin{gathered} \left(\sqrt{2}-\left(\sqrt{2}-1\right)y\right)^2+y^2=2\\ \implies\quad 2-2\sqrt{2}\left(\sqrt{2}-1\right)y+\left(\sqrt{2}-1\right)^2y^2+y^2=2\\ \implies\quad 2-\left(4-2\sqrt{2}\right)y+\left(3-2\sqrt{2}\right)y^2+y^2=2\\ \implies\quad -\left(4-2\sqrt{2}\right)y+\left(4-2\sqrt{2}\right)y^2=0\\ \implies\quad \left(4-2\sqrt{2}\right)y(-1+y)=0. \end{gathered}$ So $y=0$ (the intersection we already know about) or $y=1$.

Alternatively, substituting $y=\sqrt{2-x^2}$ into the equation of the line gives a shorter solution if we notice that $2-x^2 = \left(\sqrt{2}-x\right)\left(\sqrt{2}+x\right)$.

So the line intersects this semicircle at $\left(\sqrt{2},0\right)$ and $(1,1)$.

The sector has an angle of $45^\circ=\dfrac{\pi}{4}$. The simplest way to calculate the area of this segment is to work out the area of $\frac{1}{8}$ of the circle, and then subtract the triangle which has base $\sqrt{2}$ and height $1$.

So the area is $\frac{1}{8} \times 2\pi - \frac{1}{2} \times \sqrt{2} \times 1 = \frac{\pi}{4} - \frac{1}{\sqrt{2}}$