Geometric Interpretation of Integral

• November 15th 2009, 02:01 PM
xwrathbringerx
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, $\sqrt{a^2-x^2}$ 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.

• November 15th 2009, 02:07 PM
Jhevon
Quote:

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, $\sqrt{a^2-x^2}$ 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?
• November 15th 2009, 02:09 PM
xwrathbringerx
Ooops sori

I forgot to mention:

This is part of a question. The first part required me to find http://www.mathhelpforum.com/math-he...e428af51-1.gif between x = b and x = 0 using the substitution $u =\arcsin\frac{x}{a}$, a > b > 0.
• November 15th 2009, 02:13 PM
Veraltus
Geometric Interpretations of Integrals
-Deleted
• November 15th 2009, 02:14 PM
Jhevon
Quote:

Originally Posted by xwrathbringerx
Ooops sori

I forgot to mention:

This is part of a question. The first part required me to find http://www.mathhelpforum.com/math-he...e428af51-1.gif between x = b and x = 0 using the substitution $u =\arcsin\frac{x}{a}$, 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?
• November 15th 2009, 02:23 PM
xwrathbringerx
.... I have no idea. (Doh)
• November 15th 2009, 02:24 PM
Veraltus
Are you asking how to solve the integral?
• November 15th 2009, 02:29 PM
xwrathbringerx
The integral is just $\frac{a^2}{2}[cosb - 1]$ rite?
• November 15th 2009, 02:39 PM
Jhevon
Quote:

Originally Posted by xwrathbringerx
.... I have no idea. (Doh)

the inetgral gives the area under the semi-circle and above the x-axis between x = 0 and x = b

Quote:

Originally Posted by Veraltus
Are you asking how to solve the integral?

it does not seem that way

Quote:

Originally Posted by xwrathbringerx
The integral is just $\frac{a^2}{2}[cosb - 1]$ 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...)
• November 15th 2009, 02:43 PM
xwrathbringerx
Quote:

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?
• November 15th 2009, 02:45 PM
Jhevon
Quote:

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
• November 15th 2009, 02:50 PM
xwrathbringerx
.
• November 15th 2009, 02:56 PM
Jhevon
Quote:

Originally Posted by xwrathbringerx
.

an awkward way to ask the question. but i stand by my answer.
• November 15th 2009, 02:58 PM
xwrathbringerx
I've attached it. It's a thumbnail.

http://www.mathhelpforum.com/math-he...0-untitled.bmp
• November 15th 2009, 03:01 PM
Veraltus
Well using the standard trigonometric substitution method...and going back to our original integral i got..

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

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

so the integral becomes

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

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

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

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