Science:Math Exam Resources/Courses/MATH101/April 2017/Question 02 (a)/Solution 1

We consider the region between two curves $y^{2}=1-x$ and $2y^{2}=5-x$.

First, we find the intersection points. Since two curves can be written as $x=1-y^{2}$ and $x=5-2y^{2}$, plugging the first one into the second one, we get $$1-y^{2}=5-2y^{2}\iff y^{2}=4\iff y=\pm 2.$$ Therefore, the intersection points are $(-3,2)$ and $(-3,-2)$.

On the other hand, according to the question, the curve $x=5-2y^{2}$ lies on the right side of $x=1-y^{2}$, so that $1-y^{2}\leq 5-2y^{2}$ when $y\in (-2,2)$.

To summarize, the area between the two curves has to be an integral of the form

$$\int _{-2}^{2}|(5-2y^{2})-(1-y^{2})|\,dy=\int _{-2}^{2}(5-2y^{2})-(1-y^{2})\,dy=\int _{-2}^{2}4-y^{2}\,dy=2\int _{0}^{2}4-y^{2}\,dy.$$

The last equality follows from the fact that $4-y^{2}$ is even. Therefore, the answer is $a$.

Answer: $\color {blue}a$.