double integration for a region W which is enclosed by planes and cylinders given

**maximus101** · March 4th 2011, 02:13 AM

Evaluate
$\iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z$
where $W$
is the region enclosed by the planes $z = 0$ and $z = x + y + 5$ and by
the cylinders $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$ .

**CaptainBlack** · March 4th 2011, 02:39 AM

Originally Posted by maximus101

Evaluate
$\iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z$
where $W$
is the region enclosed by the planes $z = 0$ and $z = x + y + 5$ and by
the cylinders $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$ .

Consider changing to cylyndrical polars?

CB

**Prove It** · March 4th 2011, 02:44 AM

Originally Posted by maximus101

Evaluate
$\iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z$
where $W$
is the region enclosed by the planes $z = 0$ and $z = x + y + 5$ and by
the cylinders $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$ .

Are you sure this isn't supposed to be a triple integral?

**maximus101** · March 5th 2011, 10:25 AM

Originally Posted by CaptainBlack

Consider changing to cylyndrical polars?

CB

Hi, can you explain a bit more about how to do that please.

thanks

**maximus101** · March 5th 2011, 10:26 AM

Originally Posted by Prove It

Are you sure this isn't supposed to be a triple integral?

yes sorry it is a triple integral. I'm not sure how to integrate these things or find the limits, is it like integrating the inside one, then
integrating the answer etc.?

thank you

**DSYEAY** · March 5th 2011, 10:53 AM

Anyway, cylindrical coordinate is very much similar to polar coordinate except we have a z component thats all.

y= $r sin(\theta)$
x= $r cos(\theta)$

Rmb to include a conversion factor or (r) into the integrand.

**AllanCuz** · March 5th 2011, 10:57 AM

Originally Posted by maximus101

Evaluate
$\iint\limits_W \ x \mathrm{d}x\,\mathrm{d}y\mathrm{d}z$
where $W$
is the region enclosed by the planes $z = 0$ and $z = x + y + 5$ and by
the cylinders $x^2 + y^2 = 4$ and $x^2 + y^2 = 9$ .

Well, we can certainly switch to polar cordinates and find our bounds.

Polar cordinates are defined as,

$x=rcos \theta$
$y = rsin \theta$
$r^2 = x^2+y^2$
$dA=rdr d \theta$

Our integral is then,

$\iint_A \int_0^{z = x + y + 5 = r(sin \theta + cos \theta) + 5} (x=rcos \theta) (dA= rdr d \theta ) dz$

The above was just to show you how everything comes together. To make it a little more cleaner.

$\iint_A \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz$

Where A represents the area of our region bounded in the x-y domain. Certainly this happens between the 2 circles, that is

$x^2 + y^2 = 4 \to r^2 = 4 \to r = 2$

and

$x^2 + y^2 = 9 \to r^2 = 9 \to r = 3$

Thus,

$2 \le r \le 3$

Note that r represents the distance of the radial arm sweeping out from the origin. Think of it a circle with radius P which is centered at the origin. If we draw a line from the origin to the bounds of the circle, which would be P, we can obtain the area of the circle by rotating that radial arm, which we call R by 360 degrees. And that's what we're really going for, the area!

This topic actually requires a little more detail so I reccomend you consult your text about polar coordinates or look at the introduction to multiple integration stickied at the top of this forum!

But back to the question, we have defined our radial arm (R) but we need to sweep it across the domain to cover the entire area. In our case we have no restrictions so we are sweeping that arm from 0 to 360 degrees. We represent this by,

$\int_0^{2 \pi} d \theta$

Combining this result with what we wrote above,

$\iint_A \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz$

$\int_0^{ 2 \pi } \int_2^3 \int_0^{r(sin \theta + cos \theta) + 5} r^2 cos \theta dr d \theta dz$

Of course we can re-arrange our integral,

$\int_0^{ 2 \pi } cos \theta d \theta \int_2^3 r^2 dr \int_0^{r(sin \theta + cos \theta) + 5} dz$

You should be able to evaluate this now

**maximus101** · March 6th 2011, 02:12 PM

got it, thank you very much

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