Having Trouble with Boundaries for Triple Iterated Integrals?

[ctb94]

I am having some trouble with finding the boundaries for the first part of the problem (dz dy dx), I should be able to figure out the second part on my own. The problem is:

Set up the triple iterated integrals (using dz dy dx and d θ dr dz) to find ∫∫∫E \sqrt{x^2+y^2} dV where
E is the part of the paraboloid z = x^2 + y^2 that lies under the plane z = 4 but above the x y plane.

I start off with finding the boundaries for z and get x^2+y^2\lez\le4. This is where I become confused. I am not sure if this is correct or even how to proceed to find the other boundaries for x and y if it is.

 

[Prove It]

I am having some trouble with finding the boundaries for the first part of the problem (dz dy dx), I should be able to figure out the second part on my own. The problem is:

Set up the triple iterated integrals (using dz dy dx and d θ dr dz) to find ∫∫∫E \sqrt{x^2+y^2} dV where
E is the part of the paraboloid z = x^2 + y^2 that lies under the plane z = 4 but above the x y plane.

I start off with finding the boundaries for z and get x^2+y^2\lez\le4. This is where I become confused. I am not sure if this is correct or even how to proceed to find the other boundaries for x and y if it is.

I would start by drawing a sketch of the region (this one is easy) and also the cross-sections parallel to the x-y plane (i.e. z = 0).

Can you at least see why 0 <= z <= 4?

What shape are the cross sections? What does that suggest the best co-ordinate system to use is? What are your bounds for it?

 

[ctb94]

I believe this is a paraboloid, so I think that the cross sections would be circles. As to the coordinate system, I have to use both cartesian and cylindrical for this problem.

 

[chisigma]

I am having some trouble with finding the boundaries for the first part of the problem (dz dy dx), I should be able to figure out the second part on my own. The problem is:

Set up the triple iterated integrals (using dz dy dx and d θ dr dz) to find ∫∫∫E \sqrt{x^2+y^2} dV where
E is the part of the paraboloid z = x^2 + y^2 that lies under the plane z = 4 but above the x y plane.

I start off with finding the boundaries for z and get x^2+y^2\lez\le4. This is where I become confused. I am not sure if this is correct or even how to proceed to find the other boundaries for x and y if it is.

The use of polar coordinates is strongly reccomandable in this case...

If you were to calculate the volume would be...

$\displaystyle V = \int_{0}^{2}\ \int_{- \pi}^{\pi} \rho\ d \rho\ d \theta = 2\ \pi\ |\frac{\rho^{2}}{2}|_{0}^{2} = 4\ \pi\ (1)$

But You have an extraterm $\rho$, so that is...

$\displaystyle W = \int_{0}^{2}\ \int_{- \pi}^{\pi} \rho^{2} \ d \rho\ d \theta = 2\ \pi\ |\frac{\rho^{3}}{3}|_{0}^{2} = \frac{16}{3}\ \pi\ (2)$

Kind regards

$\chi$ $\sigma$

 

[ctb94]

That's where my problem is. It would be better to switch to polar coordinates, however, for my answer I must set up one iterated integral in terms of dz dy dx and the other iterated integral in terms of dtheta dr dz.

 

[Prove It]

The use of polar coordinates is strongly reccomandable in this case...

If you were to calculate the volume would be...

$\displaystyle V = \int_{0}^{2}\ \int_{- \pi}^{\pi} \rho\ d \rho\ d \theta = 2\ \pi\ |\frac{\rho^{2}}{2}|_{0}^{2} = 4\ \pi\ (1)$

But You have an extraterm $\rho$, so that is...

$\displaystyle W = \int_{0}^{2}\ \int_{- \pi}^{\pi} \rho^{2} \ d \rho\ d \theta = 2\ \pi\ |\frac{\rho^{3}}{3}|_{0}^{2} = \frac{16}{3}\ \pi\ (2)$

Kind regards

$\chi$ $\sigma$

Actually $\displaystyle \begin{align*} \chi \sigma \end{align*}$, the lower boundary is actually $\displaystyle \begin{align*} z = x^2 + y^2 = r^2 \end{align*}$, so that means the iterated integral should be

$\displaystyle \begin{align*} \int_0^{2\pi}{ \int_0^2{ \int_{r^2}^4{ r \,\mathrm{d}z } \,\mathrm{d}r } \,\mathrm{d}\theta } \end{align*}$

although your theta bounds don't make any difference, your integrand needs work, as when you turn it into the double integral it will be

$\displaystyle \begin{align*} \int_0^{2\pi}{ \int_0^2{ \left( 4 - r^2 \right) \, r \, \mathrm{d}r } \,\mathrm{d}\theta } \end{align*}$

 

[chisigma]

Actually $\displaystyle \begin{align*} \chi \sigma \end{align*}$, the lower boundary is actually $\displaystyle \begin{align*} z = x^2 + y^2 = r^2 \end{align*}$, so that means the iterated integral should be...

If $\displaystyle z = x^{2} +y^{2} =\rho^{2}$ and $\displaystyle 0 < z < 4$, then is $\displaystyle 0 < \rho < 2$ ...Kind reqards

$\chi$ $\sigma$