# Math Help - Double integration ftw!

1. ## Double integration ftw!

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!).

2. Hi

The volume is actually

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

3. 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?

4. 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

5. 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.

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

7. 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

8. Well done !

9. 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.

10. 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.

11. $
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.

12. Yes but be careful about the units
You cannot write $M=V$ since M is expressed in kg and V in $m^3$
Even if $\rho = 1$, its unit is $kg/m^3$