Integrals please help.

**Bert** · April 24th 2006, 03:44 AM

I have maked some mistakes in follow integrals but I can't find them.
Who can help?

$\int \int \int z \ dxdydz \ = \ \int^2_{0} \int^2_0 \int ^{\sqrt{4-y^{2}}} z \ dz dydx$

$=\int^2_0\int^2_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}} = \int^2_0\int^2_0\frac{4-y^2}{2}dydx= \int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^2$

$=\int^2_0[2y- \frac{1}{6}y^3]^2_0\ = \int^2_0\ 4 \ -\frac{8}{6} \ = \frac{8}{3}\int^2_0dx=\frac{8}{3}[x]_0^2=\frac{16}{3}$

$\int \int \int z \ dxdydz \ = \int_0^2\int_0^{2-x}\int_0^{\sqrt{4-y^2}} z \ dx dy dz$

$\int^2_0\int^{2-x}_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}}= \int^2_0\int^{2-x}_0\frac{4-y^2}{2}dydx=$ $\int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^{2-x}$

$=\int^2_0[2y- \frac{1}{6} y^3]^{2-x}_0=4$

Greets.

**topsquark** · April 24th 2006, 04:33 AM

Originally Posted by Bert

I have maked some mistakes in follow integrals but I can't find them.
Who can help?

$\int \int \int z \ dxdydz \ = \ \int^2_{0} \int^2_0 \int ^{\sqrt{4-y^{2}}} z \ dz dydx$

$=\int^2_0\int^2_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}} = \int^2_0\int^2_0\frac{4-y^2}{2}dydx= \int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^2$

$=\int^2_0[2y- \frac{1}{6}y^3]^2_0\ = \int^2_0\ 4 \ -\frac{8}{6} \ = \frac{8}{3}\int^2_0dx=\frac{8}{3}[x]_0^2=\frac{16}{3}$

$\int \int \int z \ dxdydz \ = \int_0^2\int_0^{2-x}\int_0^{\sqrt{4-y^2}} z \ dx dy dz$

$\int^2_0\int^{2-x}_0[{\frac{z^2}{2}}]_0 ^{\sqrt{4-y^{2}}}= \int^2_0\int^{2-x}_0\frac{4-y^2}{2}dydx=$ $\int^2_{0}[2y-\frac{1}{2}\ \frac{1}{3}y^3]_0^{2-x}$

$=\int^2_0[2y- \frac{1}{6} y^3]^{2-x}_0=4$

Greets.

I didn't see any real problems with the first integration, but I saw something in the second that may be giving you problems with the first as well. You are switching around the dxdydz without regarding the integrations. I'm explaining this badly, so look at the second integration problem:

$\int_0^2\int_0^{2-x}\int_0^{\sqrt{4-y^2}} z \ dx dy dz$
Note here that the limits on the x integration are 0 and $\sqrt{4-y^2}$ .

So: $\int_0^2\int_0^{2-x}\int_0^{\sqrt{4-y^2}} z \ dx dy dz$
$=\int_0^2\int_0^{2-x}\, xz|_0^{\sqrt{4-y^2}} \, dydz$
$=\int_0^2\int_0^{2-x}\, z \sqrt{4-y^2} \,dy dz$

The limits on the y integration are 0 and 2-x:
$=...=\int_0^2\, (-z/2) \left ( (x-2)\sqrt{4x-x^2} + 4 sin^{-1}(x/2-1) \right ) dz$

and the limits on the z integration are 0 and 2:
$=...=- \left ( (x-2) \sqrt{4x-x^2}+4sin^{-1}(x/2-1) \right )$

The order of the integration doesn't usually matter, but here you've got limits depending on the variables of integration, so great care must be taken in doing the integrals in the given order. ALWAYS work from the inside out. And even in the cases where you can switch the order of integration, make sure you are keeping the correct limits associated with the variables.

-Dan

**Bert** · April 24th 2006, 06:43 AM

I think that I find an easier way so forgot what i say.
You are right if you say that I change the integration steps but I think that’s not the problem.

So I go explain the complete exercise.
Compute the follow integral

$\int \int \int_{s} z \ dzdxdy$

where is the area in the first octant limited by a plane $y=0$ $z=0$
$x+y=2 \ \ 2y+x=6$ and the cylinder $y^2+z^2=4$

I think the best way to do thit is first makking a skecth

then I find the integral:

$A= \int^2_0 \int^2_{2-x} \int^{\sqrt{4y-y^2}}_0 \ z \ dz \ dy \ dx=\frac{4}{3}$

Is this oké?

Then B $B=\int^6_2 \int^{3-\frac{1}{2}x}_0 \int^{\sqrt{4-y^2}}_0 \ z \ dz \ dy \ dx =\frac{34}{3}$

Is this oké?

Then I compute and I get $\frac{4}{3} \ + \frac{34}{3} \ = \frac{38}{3}$

and this is not the same solution as in my book they get $\frac{26}{3}$

Where do I miss ? greets.

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