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!

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.

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

If this isn't review material, then talk to your instructor.

-Dan

__________________
Got a Physics question? Come on over to
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert

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

a) By definition: .

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

.

Note: Relative to the y-axis, the curve is .

__________________
There are two things you should never try to prove: the impossible and the obvious.

The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  
• 
  To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.
  

Thanks for the help!

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

__________________
Got a Physics question? Come on over to
To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.

To view links or images in signatures your post count must be 10 or greater. You currently have 0 posts.


"I must not fear. Fear is the mind killer. Fear is the little death that brings total obliteration. I will face my fear. I will permit it to pass over me and through me. And when it has gone I will turn the inner eye to see its path. Where the fear has gone there will be nothing. Only I will remain." - The Litany Against Fear, "Dune" by Frank Herbert

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.

Isn't R just equal to ? why do you need to subtract. But I can understand it if the R denotes the white area

tsal15