# Thread: finding the area of an indefinite integral

1. ## finding the area of an indefinite integral

Find the area between the parabolas y ^2 = 9x and x ^ 2 = 9y

Round the answer to two decimal places.

i think i know what to do ... solve for the points of intersection by making them equal to each other but i cant do it :S ...

than i take those points and do an indefinite integral and solve

You've got x ^ 2 = 9y => (1/9)x^2=y

Fill that into the other function, giving you

[(1/9)x^2]^2 = 9x

Solve for x to get your intersection points

x^4 / 81 = 9x

x^4 - 729x = 0

x(x-9) (x^2 +9x +81)

what do i do next ?

2. Originally Posted by rock candy
Find the area between the parabolas y ^2 = 9x and x ^ 2 = 9y

Round the answer to two decimal places.

i think i know what to do ... solve for the points of intersection by making them equal to each other but i cant do it :S ...

than i take those points and do an indefinite integral and solve
Hi

$y^2 = 9x$ and $x^2 = 9y$

$x^4 = 81y^2 = 729x$

$x(x^3 - 729) = 0$

$x(x - 9)(x^2 + 9x + 81) = 0$ which gives x=0 or x=9

3. so i used x = 0 and x= 9 ? ... what do i do next ...how do i setup my equation to integrate

4. The area is $\int \int \ dy \ dx$
You need to find the correct boundaries
Sketch a graph of both functions
You can see that the boundaries for x are 0 and 9
When x is between 0 and 9, between which values can y go ?

5. k so i have x= 0 , x= 9 ... i graphed it and found that when x = 0 , y =0 and when x=9 , y = 9 ..

so i setup my integral

Area = (integral sign) b= 9 , a=9 9y - 1/9 y^2

= 9/2 y^2 - 1/27 y^3

i sub in the points and i get

675/2 - 675/2

Area = 0

6. It is clear that for x=0, y=0 and for x=9, y=9 but for a given value for x, y cannot go from 0 to 9
Look at the graph below :for a given x, the possible values for y are given by the blue line. It means that the boundaries for y are functions of x.