Evaluate the Integral

**larryboi7** · July 30th 2010, 10:43 AM

$\int_0^3 \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} (x^2+y^2) dydx$

I dont know how to change it to cylindrical coordinates and set up the integral for it.

**yeKciM** · July 30th 2010, 11:16 AM

u can use here polar coordinates

$x= \rho cos \varphi$

$y= \rho sin \varphi$

$J= \rho$

so it will be

$\displaystyle \int_0 ^3 \int _{-\sqrt{9-x^2}} ^{\sqrt{9-x^2}} \rho^2 \rho \; d\varphi\; d\rho$

and u just need to change limits

simple

P.S here u are cylindrical coordinates

$x= \rho cos \varphi$

$y= \rho sin \varphi$

$z=z$

$J= \rho$

lol i think u need this

$\displaystyle \int_{-\frac {\pi}{2}} ^{\frac {\pi}{2}} d\varphi \int _0 ^3 \rho^3 \; d\rho$

**HallsofIvy** · July 31st 2010, 01:38 AM

First draw a picture. Set up an xy- coordinate system. Since x ranges between x= 0 and x= 3, draw vertical lines there. Now, for each x, y ranges between $y= -\sqrt{9- x^2}$ and $y= \sqrt{9- x^2}$ . But squaring either of those gives $y^2= 9- x^2$ or $x^2+ y^2= 9$ , the circle with center at (0,0) and radius 3. The right half of that circle lies between x= 0 and x= 3.

That is, the region you are integrating is a semi-circle, the portion of $x^2+ y^2= 9$ in the x> 0 half-plane.

Of course, $x^2+ y^2= r^2$ and $dxdy= r drd\theta$ .

That gives the integral yeKciM gives.
(Except that I would use "r" rather than " $\rho$ ". I reserve " $\rho$ " for spherical coordinates.)

**yeKciM** · July 31st 2010, 04:01 AM

Originally Posted by HallsofIvy

First draw a picture. Set up an xy- coordinate system. Since x ranges between x= 0 and x= 3, draw vertical lines there. Now, for each x, y ranges between $y= -\sqrt{9- x^2}$ and $y= \sqrt{9- x^2}$ . But squaring either of those gives $y^2= 9- x^2$ or $x^2+ y^2= 9$ , the circle with center at (0,0) and radius 3. The right half of that circle lies between x= 0 and x= 3.

That is, the region you are integrating is a semi-circle, the portion of $x^2+ y^2= 9$ in the x> 0 half-plane.

Of course, $x^2+ y^2= r^2$ and $dxdy= r drd\theta$ .

That gives the integral yeKciM gives.
(Except that I would use "r" rather than " $\rho$ ". I reserve " $\rho$ " for spherical coordinates.)

and looks like this when drown

yes, you are correct

but it's (for me at least) much easier to remember (or don't forget)

when using like this

Polar coordinates : $(x,y) \to (\rho , \varphi)$

$x = \rho \cos \varphi$

$y = \rho \sin \varphi$

$J= \rho$

natural limits for polar coordinates :

$\rho \ge 0$

$\displaystyle \varphi \mid _0 ^{2\pi}$

Cylindrical coordinates : $(x,y,z) \to (\rho , \varphi,z)$

$x = \rho \cos \varphi$

$y = \rho \sin \varphi$

$z = z$

$J= \rho$

natural limits for cylindrical coordinates :

$\rho \ge 0$

$\displaystyle \varphi \mid _0 ^{2\pi}$

$\displaytyle z \mid _{-\infty} ^{\infty}$

Spherical coordinates : $(x,y,z) \to (\rho ,\varphi, \theta)$

$x= \rho \cos( \varphi ) \sin (\theta)$

$y= \rho \sin( \varphi ) \sin (\theta)$

$z= \rho \cos (\theta)$

$J= \rho^2 \sin (\theta)$

natural limits for spherical coordinates:

$\rho \ge 0$

$\varphi \mid _0 ^{2\pi}$

$\theta \mid _0 ^\pi$

P.S. @ larryboi7 : when changing the limits u must always know that they can't be greater than the natural limits , but can be smaller

... let's say for spherical $\theta$ goes from $0$ to $\pi$ , and never can go to let's say from $0$ to $\frac {3\pi}{2}$ but can from $0$ to $\frac {\pi}{2}$ just for example

and if not by any condition in your task limits of $\rho, \varphi, \theta$ are changed u'll use natural limits for those

and what is more important (if u don't realize it) when for ur integral changing limits for $\rho$ it's obvious that limits are from $0$ to 3, because u have circle with center in (0,0) so ur limits goes from zero to radius of ur circle, but if ur circle is shifted (doesn't have center in point (0,0) ) then u must do a little calculating to see how do your $\rho$ changes...

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