Find the area of the region.

Question

x = 4 − y²,  x = y² − 4

Answer 1

First, we want to find where the two graphs intersect. In order to do this, we set them equal to each other. So, we get 4 - y² = y²- 4. Then, we solve the equation to find the value(s) of y. After performing algebra, we get 8=2y², so 4=y², which means y= 2 and y=-2. This means the two curves intersect when y=2 and y=-2.

In order to find the area, we have to take the integral of the equation on the right minus the equation on the left (the right would be 4-y² in this case and the left would be y²-4). If you are confused about this part, make sure to graph both equations on a graph, and you will be able to tell which one is on the right and which one is on the left. The bounds of the integral are the points where the two lines intersect, hence 2 and -2, and we are also integrating with respect to y. Now that we have all the necessary information, we can write the integral:

Area =∫²_-2 ((4-y²)-(y²-4))dy

=∫²_-2(4-y²-y²+4)dy

=∫²_-2(-2y²+8)dy

=∫²_-2(-2y²)dy+∫²_-2(8)dy

=-2(y³/3)^y=2_y=-2+8(2-(-2))

=-2(8/3+8/3)+8(2+2)

=64/3

Hence, the area is 64/3.

Answer 2

The curves intersect when y = 2 and y = -2.

Draw a diagram. The region is symmetric to the x and y axes.

Area = 2∫_{(0 to 2)} [(4 - y²) - (y² - 4))]dy = 2∫_{(0 to 2)}[8 - 2y²]dy = 4∫_{(0 to 2)} [4 - y²]dy =

4[4y - (2/3)y³]_{(0 to 2)} = 4[8 - 16/3] = 32/3