Calculus 3 help, finding the dr limits of integral for cylindrical integration

Question

wmj259 · Answer

if i have a paraboloid labeled, z= x^2+y^2 and the top z limit is z=25, im guessing the lower z limit is 0. I found the limits for dtheta as 0<theta<2pi but I cant figure out the limits for dr, Im guessing it from 0<r<5, because its a circle of radius 25? or is that only when the z is squared = x^2+y^2?

wmj259 · Answer

OR is it that I have to convert the z function in terms of r and theta?

wmj259 · Answer

so x=rcostheta and y= r sintheta and i plug that in for z=r^2? as my function into the integral?

wmj259 · Answer

$$z=25, z=x ^{2}+y ^{2}$$
thats my main function given to me.

wmj259 · Answer

i think im supposed to do $$rdzdrd \theta$$

wmj259 · Answer

so my dr lower limit is z=r^2 and upper limit is 25

empty · Answer

Yeah so far so good, keep going! :D

wmj259 · Answer

so im guessing my integrals will be $$\int\limits_{0}^{2\pi}\int\limits_{0}^{5}\int\limits_{r^{2}}^{25} (r^2)r dz dr dtheta$$

wmj259 · Answer

because its a paraboloid i thought the theta limits are all around the x-yplane so 360 degrees or 2pi. also I would think that using x=costheta and y=rsintheta to plug into the x^2+y^2=z so z=r^2. thats my lower limit for dz and upper limits 25. im lost on the radius limit, i know the lower is 0, not sure how to find the upper.

empty · Answer

Almost, but this would be like if you had a temperature distribution throughout the volume that followed $r^2$ that you wanted to integrate over the entire volume to find the total heat content. What you want is:
$$\int\limits_{0}^{2\pi}\int\limits_{0}^{5}\int\limits_{r^{2}}^{25}r dz dr d \theta$$

wmj259 · Answer

but isnt my main function that i'm supposed to integrate be z=r^2? also is my radius upper limit correct?

empty · Answer

(all I changed was I removed the $r^2$ and replaced it with a 1, since a volume of space is something that doesn't depend on where you are.)

You have the $z=r^2$ condition in your first limit of integration, so you are accounting for that.

wmj259 · Answer

howd you get the function to be integrated as basically (1)?

wmj259 · Answer

kinda looks like its just a invisible constant?

empty · Answer

Yep your radius upper limit is correct, a picture will help: |dw:1446216437534:dw|