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 $$1-y^2=5-2y^2⟺y^2=4⟺y=\pm 2.$$ 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 $1-y^2\le 5-2y^2$ when $y\in (-2,2)$.

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

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

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

Answer: $a$.