Order of Integration: How to Change from dxdydz to dydxdz?

[yazz912]

1. The problem statement, all variables and given/known
Show that

∫∫∫ 12y^2 z^3 sin[x^4] dxdydz

Region: { y< x< z
0< y< z
0 <z< (Pi)^ 1/4

Equals Pi/4
Change order of integration to dydxdz 2. Homework Equations

Order of integration
3. The Attempt at a Solution

First I graphed my region on the on the zy- plane and then I graphed my region on the xz- plane . I know my limits of integration for z will not change bc I'm still going from 0 to (Pi)^1/4

And as far as my limits for dy I have to put it in terms of x and z.
But I'm not seeing what my limits are to convert to dydxdz.

 

[BvU]

Limits are given in the formulation of the exercise. They don't change. So $$\int_0^z \, 12 y^2 dy$$ is your first to work out.

 

[yazz912]

Limits are given in the formulation of the exercise. They don't change. So $$\int_0^z \, 12 y^2 dy$$ is your first to work out.

We are instructed to change the order to dydxdz that's why.

 

[xiavatar]

Think of the z as a constant. And draw a graph of your limits as you would normally do for an area of integration in two variables in the x-y plane.

Hint*: $0\leq x\leq z, 0\leq y\leq$ ? What is the line that bounds the shaded region between the y-axis and y=x?

 

[yazz912]

Think of the z as a constant. And draw a graph of your limits as you would normally do for an area of integration in two variables in the x-y plane.
Hint*: $0\leq y\leq z, 0\leq x\leq$ ? What is the line that bounds the shaded region between the y-axis and y=x?

I would think x=y

So wrt dx limit of integration is from 0 to y?

 

[xiavatar]

Sorry I mixed up the x and y in my intervals earlier. The dx limit is from 0 to z and the dy limit is from 0 to x. Tell me if you can understand why this is so. Notice that the dx limit is clearly bounded by 0 and z based on your diagram, and the shaded area has to be bounded by 0 and the line y=x.

 

[yazz912]

Sorry I mixed up the x and y in my intervals earlier. The dx limit is from 0 to z and the dy limit is from 0 to x. Tell me if you can understand why this is so. Notice that the dx limit is clearly bounded by 0 and z based on your diagram, and the shaded area has to be bounded by and the line y=x.

Ohh ok good that's how I originally had it set but wasn't sure if it was correct... But once integrated it equals pi/4 . So it works.
Referring to z as a constant really helped out to graph the xy-plane thank you!:)

 

[BvU]

We are instructed to change the order to dydxdz that's why.

Yes. Meaning: integrate over dy, then over dx and finally over dz:
$$\int_0^{\root 4 \of \pi} \,\int_0^z \,\int_y^z \, 12 y^2 z^3 \sin x^4 \ dx dy dz = \int_0^{\root 4 \of \pi} \,\int_y^z \,\int_0^z \, 12 y^2 z^3 \sin x^4 \ dy dx dz = $$
$$ \ \ \ \int_0^{\root 4 \of \pi} \, z^3 \int_0^z \, \sin x^4\, \left ( \int_y^z \, 12 y^2 \ dy \right ) dx dz $$ etc.

 

[pasmith]

Yes. Meaning: integrate over dy, then over dx and finally over dz:
$$\int_0^{\root 4 \of \pi} \,\int_0^z \,\int_y^z \, 12 y^2 z^3 \sin x^4 \ dx dy dz = \int_0^{\root 4 \of \pi} \,\int_y^z \,\int_0^z \, 12 y^2 z^3 \sin x^4 \ dy dx dz = $$
$$ \ \ \ \int_0^{\root 4 \of \pi} \, z^3 \int_0^z \, \sin x^4\, \left ( \int_y^z \, 12 y^2 \ dy \right ) dx dz $$ etc.

That's ... not how it works.

 

[dirk_mec1]

Yes. Meaning: integrate over dy, then over dx and finally over dz:
$$\int_0^{\root 4 \of \pi} \,\int_0^z \,\int_y^z \, 12 y^2 z^3 \sin x^4 \ dx dy dz = \int_0^{\root 4 \of \pi} \,\int_y^z \,\int_0^z \, 12 y^2 z^3 \sin x^4 \ dy dx dz = $$
$$ \ \ \ \int_0^{\root 4 \of \pi} \, z^3 \int_0^z \, \sin x^4\, \left ( \int_y^z \, 12 y^2 \ dy \right ) dx dz $$ etc.

This is totally wrong!

 

[Feodalherren]

This guy did a great job in explaining it for me, maybe he can help you too

 

[LCKurtz]

@yazz912: The only way to sensibly attack a problem like this is to draw the figure in three dimensions and use it to change the order of integration. That's what you would do in 2 dimensions, isn't it? So here's a hint for this problem. I will use $h$ instead of $\sqrt[4]\pi$. On a carefully laid out xyz coordinate system mark the points $(0,0,0),(h,h,h),(h,0,h),(0,0,h)$. Join these points to form a tetrahedron. See if you can convince yourself that this solid gives the region that is being described with your original triple integral. Once you satisfy yourself that that is true, then use the figure to do the integration in the other order.

 

[BvU]

My apologies to Jazz, total blooper missing out on x > y. PA and Dirk are of course absolutely right, albeit not very constructively. Feodo is more helpful and LC gives the best advice. My blind spot must have come from too much physics ("never a problem" ) and not enough math (can be tricky ).