# Determine surface of area bounded by...

• May 1st 2008, 04:00 AM
Moo
Determine surface of area bounded by...
Hello guys !

I'd like to know the general method of doing when dealing with such texts :

"find the area bounded by y=..... and y=......"

or "find the volume of the area with a revolution around "...." axis"

In fact, it'd be nice if you could give me examples of texts and their solving generalization, pleaaaaase (Itwasntme)
I hope I'm not asking a too vague thing... :)

Thanks ^^
• May 1st 2008, 04:09 AM
wingless
If you know how to evaluate double integrals it's easy.

The area between f(x) and g(x) is

$\int_{b}^{a} \int_{g(x)}^{f(x)} ~dy~dx$

or

$\int_{d}^{c} \int_{g(y)}^{f(y)} ~dx~dy$

a and b are x coordinates, c and d are y coordinates of the intersection points of the curves.

Example:
Find the area of the circle $x^2 + y^2 = 9$.

Solution (analytic, not geometric ;p)
$x^2 + y^2 = 9$

$y^2 = 9-x^2$

This gives us two functions of x:
$f(x) = \sqrt{9-x^2}$

$g(x) = -\sqrt{9-x^2}$

So we'll find the area between these curves.

Find the x coordinates of the intersection points:

$\sqrt{9-x^2} = -\sqrt{9-x^2}$

$x = 3,-3$

So the integral is,
$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx$

$\int_{-3}^{3} 2\sqrt{9-x^2}~dx$

$x \sqrt{9-x^2}+9 \arcsin \left(\frac{x}{3}\right) ~|^{3}_{-3}$

$9\pi$
• May 1st 2008, 04:15 AM
Moo
Hi !

Quote:

Originally Posted by wingless
If you know how to evaluate double integrals it's easy.

I'm not sure to get the whole meaning of "evaluate double integrals"... I know how to calculate them that's all :P

Quote:

The area between f(x) and g(x) is

$\int_{b}^{a} \int_{g(x)}^{f(x)} ~dx~dy$

or

$\int_{d}^{c} \int_{g(y)}^{f(y)} ~dy~dx$

a and b are y coordinates, c and d are x coordinates of the intersection points of the curves.
Thanks for it, it's helping me a lot ! (Happy)

And what about things introducing multiplication by $\pi$ ? Is it always pi or can it be 2pi ? What is this about ?
• May 1st 2008, 04:35 AM
wingless
Quote:

Originally Posted by Moo
And what about things introducing multiplication by $\pi$ ? Is it always pi or can it be 2pi ? What is this about ?

Is it about the volume of a revolution formula $\pi \int^{a}_{b}f^2(x)~dx$ ?

It's always $\pi$ in this formula because it comes from the area of a circle $\pi\cdot r^2$.

If you want to find the volume of a revolution of $\alpha$ degrees, you can use it like,

$\frac{\alpha}{2\pi}\cdot\pi \int^{a}_{b}f^2(x)~dx$

$\frac{\alpha}{360}\cdot\pi \int^{a}_{b}f^2(x)~dx$
(if alpha is in degrees)
• May 1st 2008, 04:42 AM
Moo
Oh ! (Clapping)
And how are defined a and b ? Intersection of the curve with the axis ?
I didn't look further, but is there an importance in the order of a and b ?
• May 1st 2008, 04:47 AM
wingless
Quote:

Originally Posted by Moo
Oh ! (Clapping)
And how are defined a and b ? Intersection of the curve with the axis ?
I didn't look further, but is there an importance in the order of a and b ?

No, intersection of the curves f(x) and g(x). Set f(x) = g(x) and you'll get two x values, these are a and b. Normally, b must be the least one and a must be the bigger one. But it doesn't matter if you change their order, you'll get the negative of the same value.

$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx = 9\pi$

$\int_{3}^{-3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx = -9\pi$

Also, as a general rule,

$\int_{b}^{a}f(x)~dx = - \int_{a}^{b}f(x)~dx$
• May 1st 2008, 04:55 AM
Moo
I'm gonna be lectured because I didn't quote... :(

Quote:

Originally Posted by wingless
Is it about the volume of a revolution formula $\pi \int^{a}_{b}f^2(x)~dx$ ?

• May 1st 2008, 05:06 AM
wingless
Sorry, I don't have time to explain them at the moment, but these sites explain it as well:

http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node22.html
Volume of A Solid of Revolution
• May 1st 2008, 05:16 AM
Moo
Hey, thanks ! This is really enough for me and I think I understand where those weird formulas come from :D

(Bow)
• May 1st 2008, 07:44 AM
Krizalid
Quote:

Originally Posted by wingless

So the integral is,
$\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} ~dy~dx$

It's actually easier by transforming this into a polar transform. So, $\int_{-3}^{3}{\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}}{dy}\,dx}=\int_{0}^{2\pi }\!{\int_{0}^{3}{r\,dr}\,d\theta }=9\pi .$

Or by considering $\int_{-3}^{3}{\sqrt{9-x^{2}}\,dx},$ which we're computing the area of a semicirle whose radius is $3.$
• May 1st 2008, 08:19 AM
Moo
It's not the problem of transforming into polar coordinates or finding tricks, I wanted the method and I have to admit, it was clear with what wingless did :D

Quote:

which we're computing the area of a semicirle whose radius is 3
Quote:

Solution (analytic, not geometric ;p)
• May 1st 2008, 08:21 AM
Krizalid
Sure. I read the last quote too. I posted just in case. :)
• May 1st 2008, 08:24 AM
Moo
Quote:

Originally Posted by Krizalid
Sure. I read the last quote too. I posted just in case. :)

It's ok for that, thanks :)

I really needed elements to understand problems that are sometimes posted here (Giggle)

Do you know how to do when the axis around which there is the revolution is different from y=0 ?