# finding the area between two curves

• Apr 18th 2010, 03:07 PM
rubbersoul923
finding the area between two curves
Area between two curves:
y=9-x^2
y=x^2 - 7

Show that the area is (128√2)/3

So...
y=9-x^2
y'=-2x
I(x) = 9x - (1/3)x^3

y=x^2 - 7
y'=2x
I(x) = (1/3)x^3 - 7x

9 - x^2 = x^2 - 7
x = √8

Correct so far? What next? How can I make it to find the correct area? Thanks!
• Apr 18th 2010, 03:14 PM
Debsta
Quote:

Originally Posted by rubbersoul923
Area between two curves:
y=9-x^2
y=x^2 - 7

Show that the area is (128√2)/3

So...
y=9-x^2
y'=-2x
I(x) = 9x - (1/3)x^3

y=x^2 - 7
y'=2x
I(x) = (1/3)x^3 - 7x

9 - x^2 = x^2 - 7
x = √8

Correct so far? What next? How can I make it to find the correct area? Thanks!

x = =/- root 8
These are then the limits for the integration.
• Apr 18th 2010, 03:25 PM
rubbersoul923
So then the -√8 and √8...

I plug them into each integral
I(x) = 9x - (1/3)x^3
I(√8) = 9(√8) - (1/3)(√8)^3 = 17.9
I(-√8) = 9(-√8) - (1/3)(-√8)^3 = -17.9

I(x) = (1/3)x^3 - 7x
I(√8) = (1/3)(√8)^3 - 7(√8) = -12.2
I(-√8) = (1/3)(-√8)^3 - 7(-√8) = 12.2

How do I arrive at the area of 128√2 / 3 ?

Area = b-a
Area = which one is b and which is a?
• Apr 18th 2010, 03:29 PM
Debsta
Quote:

Originally Posted by rubbersoul923
So then the -√8 and √8...

I plug them into each integral
I(x) = 9x - (1/3)x^3
I(√8) = 9(√8) - (1/3)(√8)^3 = 17.9
I(-√8) = 9(-√8) - (1/3)(-√8)^3 = -17.9

I(x) = (1/3)x^3 - 7x
I(√8) = (1/3)(√8)^3 - 7(√8) = -12.2
I(-√8) = (1/3)(-√8)^3 - 7(-√8) = 12.2

How do I arrive at the area of 128√2 / 3 ?

Area = b-a
Area = which one is b and which is a?

Draw a quick sketch. y = 9-x^2 is an "upside-down parabola" with y -int 9 and axis of symmetry the y-axis. Also draw y = x^2 -7 (you should know what it looks like). So you want to calculate the integral from -rt8 to +rt8 of the function ((9-x^2) - (x^2 -7))