double integration questions help please

**tsang** · Sep 9th 2010, 07:03 AM

Hi everyone, can anybody please show me how to solve these two double integration questions please?
Here they are:

(a) $\int_0^1 \int_0^{\arccos(x)} \cos(\sin y)} \, dy \,dx<br />$
(b) $\int_0^4 \int_{\sqrt{2}}^{3} \int_0^y z\cos (y^6) \, dx \,dy \,dz$

For the first one is painful,I cannot even start,but I guess maybe I need to change the order,am I right?
Second one is a bit horrible, I cannot continue after I integrated with respect to x, it was getting really messy.
Please show me what I should do,thanks a lot.

**yeKciM** · Sep 9th 2010, 07:15 AM

are you sure that you write that correctly ?

$\cos ({\sin {y}})$

**Ackbeet** · Sep 9th 2010, 07:27 AM

For the first problem, change the order of integration. Look at the region over which you're integrating and carefully look at the new limits. The result will be much easier to work with.

**Ackbeet** · Sep 9th 2010, 07:42 AM

For the second integral, are you sure it isn't

$\displaystyle{\int_{0}^{4} \int_{\sqrt{2}}^{3} \int_{0}^{y}z\cos^{6}(y) \, dx \,dy \,dz?}$

The one you wrote is nasty and has incomplete gamma functions everywhere.

**tsang** · Sep 9th 2010, 05:38 PM

Originally Posted by Ackbeet

For the first problem, change the order of integration. Look at the region over which you're integrating and carefully look at the new limits. The result will be much easier to work with.

Thank you,I will try to change the order,and see if it works.

**tsang** · Sep 9th 2010, 05:40 PM

Originally Posted by Ackbeet

For the second integral, are you sure it isn't

$\displaystyle{\int_{0}^{4} \int_{\sqrt{2}}^{3} \int_{0}^{y}z\cos^{6}(y) \, dx \,dy \,dz?}$

The one you wrote is nasty and has incomplete gamma functions everywhere.

Yes, friend, I'm sure I wrote the right thing. I know this one is very nasty because of that $y^6$ , that's why I found it is so hard. Can you please help me with that? Thanks a lot.

**Ackbeet** · Sep 10th 2010, 02:11 AM

With regard to the second integral, here's what Mathematica gives:

$\displaystyle{\int_{0}^{4}\int_{\sqrt{2}}^{3}\int_ {0}^{y}z\,\cos(y^{6})\,dx\,dy\,dz=}$
$\displaystyle{-\frac{2}{3}(-1)^{1/6}\left(-\Gamma\left(\frac{1}{3},-8i\right)+(-1)^{2/3}\Gamma\left(\frac{1}{3},8i\right)+\Gamma\left(\f rac{1}{3},-729i\right)-(-1)^{2/3}\Gamma\left(\frac{1}{3},729i\right)\right).}$

Here the $\Gamma$ is the upper incomplete Gamma function defined by

$\displaystyle{\Gamma(s,x)=\int_{x}^{\infty}t^{s-1}e^{-t}\,dt.}$

The numerical value of this result is approximately $-0.326074.$

I might be able to explain portions of the journey to get you there.

Let's take your original integral. You can immediately do the z integration. You have the following:

$\displaystyle{\int_{0}^{4}\int_{\sqrt{2}}^{3}\int_ {0}^{y}z\,\cos(y^{6})\,dx\,dy\,dz=\left(\frac{z^{2 }}{2}\right)\Bigg|_{0}^{4}\int_{\sqrt{2}}^{3}\int_ {0}^{y}\cos(y^{6})\,dx\,dy=8\int_{\sqrt{2}}^{3}\in t_{0}^{y}\cos(y^{6})\,dx\,dy.}$

Next, we perform the $x$ integral to obtain

$\displaystyle{8\int_{\sqrt{2}}^{3}y\,\cos(y^{6})\, dy.}$

From here, I'd probably do the substitution $u=y^{2},$ from which we get $du=2y\,dy.$ Hence, the integral becomes

$\displaystyle{4\int_{2}^{9}\cos(u^{3})\,du.}$

This function actually has an antiderivative in terms of the gamma function. I'm not sure I can explain it to you. Perhaps Opalg could weigh in on that one. Once you have that antiderivative, all you do is plug in the limits, and you have your result.

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