# Thread: Area enclosed by several functions

1. ## Area enclosed by several functions

Hey

I don't understand how to find the area enclosed by functions. I tried my text book, but it is pretty confusing. Can someone show me the steps to this problem, please?

*R is closed by y= square root of x , x=9, and y=0

a) Integrate with respect to the x -axis

b) Integrate with respect to the y-axis.

Thanks!

2. If you draw it out, you'll find that you're basically finding the area under the function from x = 0 to x = 9. So it's just a simple definite integral.

3. Uh, yeah, I still don't get what that means ;p

4. Originally Posted by NoiseandAttack
Uh, yeah, I still don't get what that means ;p
If this isn't review material, then talk to your instructor.

-Dan

5. Originally Posted by NoiseandAttack
Hey

I don't understand how to find the area enclosed by functions. I tried my text book, but it is pretty confusing. Can someone show me the steps to this problem, please?

*R is closed by y= square root of x , x=9, and y=0

a) Integrate with respect to the x -axis

b) Integrate with respect to the y-axis.

Thanks!
A sketch graph of the area showing all important features is essential.

a) By definition: $A = \int_{0}^{9} \sqrt{x} \, dx = \int_{0}^{9} x^{1/2} \, dx = ......$.

b) The area is the rectangular area (3)(9) - area between curve and y-axis:

$A = (3)(9) - \int_{0}^{3} y^2 \, dy = 27 - \int_{0}^{3} y^2 \, dy = ......$.

Note: Relative to the y-axis, the curve is $x = y^2$.

6. Thanks for the help!

7. Why do you do (3)(9) minus the integral from 0 to 3 of y^2?

8. Originally Posted by NoiseandAttack
Why do you do (3)(9) minus the integral from 0 to 3 of y^2?
The integral is the area between the curve and the y axis in this case, which is the "opposite" of the area you are interested in. You want the area of the 3 x 9 box less the area represented by the integral.

-Dan

9. An illustration:

Notice the rectangle formed by the y-axis, the x-axis, x = 9, and y = 3 as mentioned by mr fantastic. So R is the area of the rectangle minus the white area.

10. Originally Posted by o_O
So R is the area of the rectangle minus the white area.
Isn't R just equal to $\int_{0}^{9} \sqrt{x} dx$ ? why do you need to subtract. But I can understand it if the R denotes the white area

tsal15