# Thread: Very easy definite integral question

1. ## 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.

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?

2. 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.

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 ...

3. $y = \sqrt{9-x^2} \Rightarrow y^2 = 9 - x^2 \Rightarrow x^2 + y^2 = 9, \ y \geq 0$

So what is the area?

4. Would that just be half of the radius of the circle with radius 3?

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

6. Originally Posted by ninjaku
Would that just be half of the radius of the circle with radius 3?
Look at skeeter's graph again..

7. 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.