# Triple Integral Spherical Coordinates?

I don't think so since it's not a sphere (disk). I have not learned about cylindrical coordinates and Cartesian is just a pain, so I am assuming I am supposed to use polar or something.

Can someone clear up my confusion?

$$\int\int\int_E y\,dV$$

where E lies above the plane z=0, under the plane z=y and inside the cylinder $x^2+y^2=4$

How would you proceed here if you DON'T want to use Cartesian Coordinates?

Casey

Duuuudddddeeeeeeee........... I don't get it.... I see a cirlcle (the domain of E) and I automatically think polar... but it is 3-D so tha doesn't male sense right? For 3-D you need spherical coordinates, but that doesn't make sense either.

I am thinking I am not supposed to use Cartesians, cause that integrals a pita. But maybe I am wrong; however we usually don't get integrals that require the use of tables (for now).

Dick
Homework Helper
To find the volume you integrate the height y, over the area in the x-y plane. Use polar coordinates in the x-y plane. That's really cylindrical coordinates, but you don't have to call it that.

Hey Dick! I don't have to much experience with this. Do I set up the integral like this:

$$\int\int_D\int_0^y y\,dz\,dA$$

where D is the region bounded by x=0 and x^2+y^2=4

thus the integral becomes

$$\int_{\theta=\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{r=0}^2\,y^2\,dr\,d\theta$$

and then just plug in what y is in polar coordinates? Or did I F this one up royally?

Thanks Last edited:
Dick
Homework Helper
Not that royally, but it could use some work. Why y^2? If you are integrating y over the area in the x-y plane it's only a double integral over r and theta. And doesn't theta go from 0 to pi? And isn't the area measure in polar coordinates r*dr*dtheta?

I am missing something... where did the third integral go then? The y^2 came from integrating y*dz wtr z--->y*z from z=0 to z=y.

Then I though you integrate that over the x-y area?

Dick
Homework Helper
If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.

Maybe I should start over here. If I take the Volume and project it onto the x-y plane, I get D, which takes the form of a semi-circle that follows a path from theta=0 to pi at a radial distance 2 from the origin.

Now I want to integrate the function y wrt r and theta? Is that what you are saying?

Dick
Homework Helper
Yes, that's exactly what I'm saying. You integrate the height of the region over the area of the base.

If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.

Why am I integrating 1 ? I am looking for its 4-D volume (or whatever it is called) aren't I.

If you are setting this up as a triple integral, you integrate 1 over the coordinate limits to get the volume. The integral of 1*dz from z=0 to z=y is y.

$$\int\int\int y\,dz\,dr\,d\theta$$

Once more. Is this my integral? I am just confused, now.

I think it's time for bed! Thanks for the help Dick I will look at it in the a.m. I am getting too flustered right now. I understand every other triple integral assigned to us except this one. Moreover, I can solve this one in Cartesian Coordinates just fine, so long as I have an integral table handy. So I have no idea why this does not make sense to me in Cylindricals.

In Cartesian I have

$$\int_x\int_y\int_z\,y\,dz\,dy\,dx$$ .... Not sure what the big deal is now.

Dick
Homework Helper
Why am I integrating 1 ? I am looking for its 4-D volume (or whatever it is called) aren't I.

I'm going to pretend you didn't say that. A volume integral is the integral of 1 over the 3d coordinate limits. Look it up.

I'm going to pretend you didn't say that. A volume integral is the integral of 1 over the 3d coordinate limits. Look it up.

What is the definition of the $$\int\int\int_E f(x,y,z)\,dV$$

It is not the volume of E

correct? If I wanted the volume of E I would integrate the function

$$\int\int\int_E \,dV$$

Correct?

But my problem does not ask me to find the volume of E. It asks me to evaluate

$$\int\int\int_E f(x,y,z)\,dV$$

where f(x,y,z) is simply y. Is that not the case here? I am NOT looking for volume (in the three dimensional case)?

Dick
Homework Helper
What is the definition of the $$\int\int\int_E f(x,y,z)\,dV$$

It is not the volume of E

correct? If I wanted the volume of E I would integrate the function

$$\int\int\int_E \,dV$$

Correct?

But my problem does not ask me to find the volume of E. It asks me to evaluate

$$\int\int\int_E f(x,y,z)\,dV$$

where f(x,y,z) is simply y. Is that not the case here? I am NOT looking for volume (in the three dimensional case)?

Yes. Completely, utterly, right. I somehow thought you were looking for the volume and you're not. Sorry, sorry. Yes, the y^2 is correct. But shouldn't it be y^2/2?

I don't think so: if I have $$\int\int_D\int_0^y y\,dz\,dA$$

Just taking care of that innermost integral with respect to z I have

$$\int_0^yy\,dz=y\int_0^y\,dz=y*z|_0^y=y^2$$

yes?

Dick
Homework Helper
Yep, you win again. For a fixed x,y y is a constant. You might be better at this than I am.

Yep, you win again. For a fixed x,y y is a constant. You might be better at this than I am.

Yessss! I win! Just kidding. Okay so back to bed. I think I'll be able to sleep now So back to my proposal:

Now I have integrated the function with respect to height and got y^2. Now I evaluate a double integral over D of y^2 (but in polar terms=(r*sintheta)^2). Something to the affect of:

$$\int\int_D\,y^2\,dA=\int_0^\pi \int_0^2\, r^2*sin^2\theta\,* r\,dr\,d\theta$$

Dick