Double integration ftw!

**Showcase_22** · Apr 7th 2009, 08:37 AM

Find the volume of the solid in the first octant bounded by the coordinate planes, the plane $x=3$ , and the parabolic cylinder $z=4-y^2$ .

I've drawn a nice picture (essential) and I here's what i've deduced:

$\int_0^3 \int_0^x 4-y^2 \ dy \ dx$

$\int_0^3 \left[4y-\frac{y^3}{3}\right]_0^x$

$\int_0^3 4x-\frac{x^3}{3}dx$

$\left[2x^2-\frac{x^4}{12}\right]^3_0$

$9\left(2-\frac{9}{12}\right)$

$=\frac{45}{4}$

However, there's something incorrect here since the correct answer is 16.

Can anyone see a mistake anywhere? (especially with my limits!).

**running-gag** · Apr 7th 2009, 08:48 AM

Hi

The volume is actually

$\int_0^3 \int_0^2 \int_0^{4-y^2}\ dz \ dy \ dx = \int_0^3 \int_0^2 (4-y^2) \ dy \ dx$

**Showcase_22** · Apr 7th 2009, 09:12 AM

But doesn't that imply that it's a square region on the x-y plane?

When I drew the diagram I got this:

I drew red arrows on it to show my reasoning for the limits.

Why is this reasoning wrong?

**running-gag** · Apr 7th 2009, 09:22 AM

Originally Posted by Showcase_22

But doesn't that imply that it's a square region on the x-y plane?

Yes it is actually a rectangular region
For any value for x between 0 and 3, y can vary between 0 and 2
y is not linked to x

**Showcase_22** · Apr 7th 2009, 09:34 AM

ah, I think I get it.

$z=4-y^2$

But $z=0 \Rightarrow$ $4-y^2=0 \Rightarrow y^2=4 \Rightarrow y=\pm 2$

I see where i'm wrong as well. I thought that an octant was $\mathbb{R}^2$ split into 8 sections. A google search has revealed my foolishness!

Thankyou, I understand now.

**running-gag** · Apr 7th 2009, 11:53 AM

Originally Posted by Showcase_22

ah, I think I get it.

$z=4-y^2$

But $z=0 \Rightarrow$ $4-y^2=0 \Rightarrow y^2=4 \Rightarrow y=\pm 2$

I see where i'm wrong as well. I thought that an octant was $\mathbb{R}^2$ split into 8 sections. A google search has revealed my foolishness!

Thankyou, I understand now.

OK
It is indeed $\mathbb{R}^3$ split into 8 sections

: $x \geq 0 ; y \geq 0 ; z \geq 0$

**DeMath** · Apr 7th 2009, 11:55 AM

Find the volume of the solid in the first octant bounded by the coordinate planes, the plane $x=3$ and the parabolic cylinder $z=4-y^2$ .

See this picture carefully

**running-gag** · Apr 7th 2009, 12:13 PM

Well done !

**Showcase_22** · Apr 8th 2009, 01:31 AM

There's one more thing I don't quite understand.

$ \int_0^3 \int_0^2 \int_0^{4-y^2}\ dz \ dy \ dx = \int_0^3 \int_0^2 (4-y^2) \ dy \ dx $

I evaluated the integrand w.r.t z on the left and you do indeed get what's written on the right. However, geometrically, what does it mean when we integrate 1 w.r.t z?

To me it seems to suggest that it's the volume of a unit cube, but the whole point of triple integrals is that the volume of these cubes is made really small (ie. side length $\rightarrow \ 0$ ). If that is the case, I would expect to be evaluating a function of a variable w.r.t z and not just 1.

**running-gag** · Apr 8th 2009, 02:06 AM

In Cartesian coordinates the elementary volume unit is dx dy dz.
To get the volume you need to integrate dx dy dz over the boundaries.

Lets's go to physics. When a solid has a non-uniform density $\rho(x,y,z)$ you can evaluate its mass through $M = \int \int \int \rho(x,y,z) \ dx \ dy \ dz$

When the density is uniform $\rho(x,y,z) = \rho_0$ then the mass is $M = \int \int \int \rho_0 \ dx \ dy \ dz = \rho_0 \int \int \int \ dx \ dy \ dz = \rho_0 V$ where V is the volume of the solid.

I hope this answers your question

**Showcase_22** · Apr 8th 2009, 04:06 AM

$ M = \int \int \int \rho(x,y,z) \ dx \ dy \ dz $

But $\rho=\frac{M}{V}$ where $V$ is the volume.

Therefore if we let $\rho=1 \Rightarrow M=V$ .
$\Rightarrow \rho(x,y,x)=1$ .

This gives:

$ M=V= \int \int \int 1 \ dx \ dy \ dz $

which is a formula to find the volume.

Thanks! I get it now.