Integrals (Area of the region enclosed)

Question

Use horizontal strips to find the area, that is, integrate with respect to y.

First find the y coordinates of the two points where y = 4.5x meets x = 8-y^2.

lower limit c = ___

and

upper limit d = ___

To find the area of the enclosed region from c to d we will integrate: from c to d, g(y)dy where g(y) = ___

Evaluate the definite integral to find Area = ___

Answer 1

x=8-(4.5x)^2=8-20.25x^2

20.25x^2+x-8=0

x=(-1+/-sqr649)/40.5

=about -.654 or .604

= -.7 or .6

y=4.5(-.7) or 4.5(.6)=

= -2.94 or 2.72

y=4.5x=9x/2

x=2y/9=8-y^2

2y=72-9y^2

9y^2-2y-72=0

y=-2/18 +/-(1/18)sqr(4+36(72))

=about (-2+/-50.951)/18

=-2.942 or 2.7195

upper y limit =about 2.7

lower y limit= about -2.9

corresponding x values

x=2y/9=5.884/9 or -5.439/9

= -0.654 or 0.604

the intersection points are about (-.7, -2.9) and (.6, 2.7)

when you integrate g(y) use two intervals (-.7,-2.9) to (0,0)

and (0.0) to (.6, 2.7)

then add those two areas