Sec 7.1 prob 5 Sketch the region enclosed by the curves x=5y^2 and x=y^2+4

Question

List the points of intersection of these curves from bottom to top in the form (x,y):

(5,__) and (5,1) ***II couldn't figure of the second part to the first set

Decide whether to integrate with respect to x or y to find the area A of this region. A=?

Answer 1

The intersection points are (5,1) and (5,-1) because

5y^2 = y^2 + 4

4y^2 = 4

y^2 = 1

y = +or- 1

In either case, x=5, as 5(1)^2 = 5 and 5(-1)^2 = 5 as does (1)^2 +4 = 5 and (-1)^2 +4 = 5

y^2+4 is further to the right .

So integral ( y^2 +4 - 5y^2) = integral ( 5 - 5y^2)

= 5y - (5/3)y^3

from y = -1 to 1

(5 - 5/3) - (-5 - (-5/3)) =

(15/3 - 5/3) - ( -15/3 + 5/3) =

10/3 - (-10/3) = 20/3 = 6 and 2/3