# How do you find the area between y^2=-4(x-1) and y^2=-2(x-2)?

May 21, 2018

#### Explanation:

Here is a picture of the region whose area we seek.

When we first learn to find areas by integration, we take representative rectangles vertically.
The rectangles have base $\mathrm{dx}$ (a small change in $x$) and heights equal to the greater $y$ (the one on upper curve) minus the lesser $y$ value (the one on the lower curve). We then integrate from the smallest $x$ value to the greatest $x$ value.

That approach is quite challenging in this problem.
Here it is very valuable to learn to reflect our thinking ${90}^{\circ}$.

We will take representative rectangles horiontally.
The rectangles have height $\mathrm{dy}$ (a small change in $y$) and bases equal to the greater $x$ (the one on rightmost curve) minus the lesser $x$ value (the one on the leftmost curve). We then integrate from the smallest $y$ value to the greatest $y$ value.

Notice the duality

{:("vertical ", iff ," horizontal"), (dx, iff, dy), ("upper", iff, "rightmost"), ("lower", iff, "leftmost"), (x, iff, y):}

The phrase "from the smallest $x$ value to the greatest $x$ value." indicates that we integrate left to right. (In the direction of increasing $x$ values.)

The phrase "from the smallest $y$ value to the greatest $y$ value." indicates that we integrate bottom to top. (In the direction of increasing $y$ values.)

Here is a picture of the region with a small rectangle indicated:

The $x$ on the right (the greater $x$ value) lies on the graph of ${y}^{2} = - 2 \left(x - 2\right)$.
Solving for $x$, we get ${x}_{\text{right}} = - {y}^{2} / 2 + 2$

The $x$ on the left (the lesser $x$ value) lies on the graph of ${y}^{2} = - 4 \left(x - 1\right)$.
Solving for $x$, we get ${x}_{\text{left}} = - {y}^{2} / 4 + 1$

$y$ varies from $- 2$ to $2$, so the area of the region is

${\int}_{-} {2}^{2} \left(\left(- {y}^{2} / 2 + 2\right) - \left(- {y}^{2} / 4 + 1\right)\right) \mathrm{dy}$

$= {\int}_{-} {2}^{2} \left(1 - {y}^{2} / 4\right) \mathrm{dy} = \frac{8}{3}$