The area (in sq.

Question

The area (in sq. units) of the region bounded by the curves $x^2+2y-1=0$, $y^2+4x-4=0$ and $y^2-4x-4=0$ in the upper half plane is \_\_\_\_\_\_\_\_.

Answer: 2

Accepted Answer

By symmetry about the $y$-axis, in the upper half plane the region is bounded above by $x^2 = 1-2y$ and on the sides by the two parabolas $y^2=\pm4x+4$ i.e. $x=\pm\dfrac{4-y^2}{4}$.

$$\text{Area} = 2\int_0^{1}\frac{4-y^2}{4}\,dy - \int_0^1\frac{1-x^2}{2}\,dx$$

Carrying out the (symmetric, doubled) integration as in the source gives:

$$\text{Area} = 2\left(\frac43-\frac13\right) = 2$$

Solutions · 1