# Very easy definite integral question

• Apr 5th 2010, 03:07 PM
Archduke01
Very easy definite integral question
Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.
http://euler.vaniercollege.qc.ca/web...53ef350ee1.png

I don't know why, but I'm messing this up. I do x = 3 (so y = 0) and x = -3 (so $y = 3 \sqrt 2$) to get a graph that looks like a triangle. Using simply geometry to find the answer, I do 6 (the base) * $3 \sqrt 2$ (the height) divided by 2 to get $9 \sqrt 2$. Which is wrong.

But why?
• Apr 5th 2010, 03:11 PM
skeeter
Quote:

Originally Posted by Archduke01
Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.
http://euler.vaniercollege.qc.ca/web...53ef350ee1.png

I don't know why, but I'm messing this up. I do x = 3 (so y = 0) and x = -3 (so $y = 3 \sqrt 2$) to get a graph that looks like a triangle. Using simply geometry to find the answer, I do 6 (the base) * $3 \sqrt 2$ (the height) divided by 2 to get $9 \sqrt 2$. Which is wrong.

But why?

the graph is not a triangle ...
• Apr 5th 2010, 03:12 PM
Defunkt
$y = \sqrt{9-x^2} \Rightarrow y^2 = 9 - x^2 \Rightarrow x^2 + y^2 = 9, \ y \geq 0$

So what is the area?
• Apr 5th 2010, 03:26 PM
ninjaku
Would that just be half of the radius of the circle with radius 3?
• Apr 5th 2010, 03:29 PM
skeeter
Quote:

Originally Posted by ninjaku
Would that just be half of the radius of the circle with radius 3?

huh?
• Apr 5th 2010, 03:36 PM
Defunkt
Quote:

Originally Posted by ninjaku
Would that just be half of the radius of the circle with radius 3?

Look at skeeter's graph again..
• Apr 6th 2010, 04:37 AM
HallsofIvy
Quote:

Originally Posted by ninjaku
Would that just be half of the radius of the circle with radius 3?

I presume you mean half the area of the circle with radius 3.