Thread: Geometric Interpretation of Integral

1. Geometric Interpretation of Integral

Could someone please explain to me what it means when I have to give a geometric interpretation of each term of the integral, say, $\displaystyle \sqrt{a^2-x^2}$ between x = b and x = 0?

From what I can make of the question, all I can think of is that a is the radius of the semicircle.

2. Originally Posted by xwrathbringerx
Could someone please explain to me what it means when I have to give a geometric interpretation of each term of the integral, say, $\displaystyle \sqrt{a^2-x^2}$ between x = b and x = 0?

From what I can make of the question, all I can think of is that a is the radius of the semicircle.

Are you sure it is not x = a, as opposed to x = b? What is b?

3. Ooops sori

I forgot to mention:

This is part of a question. The first part required me to find between x = b and x = 0 using the substitution $\displaystyle u =\arcsin\frac{x}{a}$, a > b > 0.

4. Geometric Interpretations of Integrals

-Deleted

5. Originally Posted by xwrathbringerx
Ooops sori

I forgot to mention:

This is part of a question. The first part required me to find between x = b and x = 0 using the substitution $\displaystyle u =\arcsin\frac{x}{a}$, a > b > 0.
Ok, well, we know that we can interpret the integral, geometrically, to mean the net area under the curve... what can you do with this?

6. .... I have no idea.

7. Are you asking how to solve the integral?

8. The integral is just $\displaystyle \frac{a^2}{2}[cosb - 1]$ rite?

9. Originally Posted by xwrathbringerx
.... I have no idea.
the inetgral gives the area under the semi-circle and above the x-axis between x = 0 and x = b

Originally Posted by Veraltus
Are you asking how to solve the integral?
it does not seem that way

Originally Posted by xwrathbringerx
The integral is just $\displaystyle \frac{a^2}{2}[cosb - 1]$ right?
actually, no. if you differentiate that, you do not get the original integrand...

why do you not follow the hint? (write it as x = asin(u) though...)

10. the inetgral gives the area under the semi-circle and above the x-axis between x = 0 and x = b
Hmmmm but does that cover "each term of the integral" since the question asked for the geometric interpretation of all the terms?

11. Originally Posted by xwrathbringerx
Hmmmm but does that cover "each term of the integral" since the question asked for the geometric interpretation of all the terms?
will you please post the question EXACTLY as it was stated, giving any preliminary info that may have been there also

12. .

13. Originally Posted by xwrathbringerx
.
an awkward way to ask the question. but i stand by my answer.

14. I've attached it. It's a thumbnail.

15. Well using the standard trigonometric substitution method...and going back to our original integral i got..

$\displaystyle \int \sqrt{a^2 -x^2}dx$from x = 0 to x = b...

so use the substitution $\displaystyle x=asin(\theta)$ and $\displaystyle dx = acos(\theta)d\theta$

so the integral becomes

$\displaystyle \int \sqrt{a^2 - a^2 sin^2(\theta)}acos(\theta)d\theta$ which becomes $\displaystyle \int \sqrt{a^2 (1-sin^2 (\theta))}acos(\theta)d\theta$

so we finally get $\displaystyle \int acos^2(\theta)d\theta$

We have to also change our limits of integration... from x = 0 to $\displaystyle \theta$ = 0 and we get that x = b turns to $\displaystyle \theta$ = $\displaystyle arcsin(\frac {b}{a})$

so my final answer was... $\displaystyle \frac {a^2sin^{-1}( \frac {b}{a})}{2} + \frac {a^2 sin (2*sin^{-1} (\frac {b}{a}))}{4}$