Here is a picture of the region whose area we seek.
![enter image source here]()
When we first learn to find areas by integration, we take representative rectangles vertically.
The rectangles have base dxdx (a small change in xx) and heights equal to the greater yy (the one on upper curve) minus the lesser yy value (the one on the lower curve). We then integrate from the smallest xx value to the greatest xx value.
That approach is quite challenging in this problem.
Here it is very valuable to learn to reflect our thinking 90^@90∘.
We will take representative rectangles horiontally.
The rectangles have height dydy (a small change in yy) and bases equal to the greater xx (the one on rightmost curve) minus the lesser xx value (the one on the leftmost curve). We then integrate from the smallest yy value to the greatest yy 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:
![enter image source here]()
The x on the right (the greater x value) lies on the graph of y^2=-2(x-2).
Solving for x, we get x_"right" = -y^2/2+2
The x on the left (the lesser x value) lies on the graph of y^2=-4(x-1).
Solving for x, we get x_"left" = -y^2/4+1
y varies from -2 to 2, so the area of the region is
int_-2^2 ((-y^2/2+2)-(-y^2/4+1)) dy
= int_-2^2 (1-y^2/4) dy = 8/3
