Question

calculus question. Totally have no idea what Im doing... well actually I do but I suck. calculus question. Totally have no idea what Im doing... well actually I do but I suck. @Mathematics

Answer 1

gimmie a second to type this up

Answer 2

\[\int\limits_{0}^{\pi \over 2}\int\limits_{0}^{3}\int\limits_{0}^{e ^{-r^2}}r \space dz \space dr \space d \theta\]sketch the the solid region whose volume is given by the iterated integral and evaluate the iterated integral

Answer 3

I know I let u = -r^2 and that is about it lol

Answer 4

i think I can also let du=-2rdr

Answer 5

okay after this I have no clue. And Im not even sure if my substitution for u is right in the first place >.>

Answer 6

always sketch it first

Answer 7

the integral all the way to the right corelates to the first d?? we see at the end

Answer 8

that means the first integral evaulates from 0 to e^-rz with respect to z

Answer 9

yes

Answer 10

if you evaulaute the first z integral you get er-r^2

Answer 11

i believe this is the function that represent the height of your cylinder

Answer 12

now we can look at the bounds of r and thet to find the overall shape

Answer 13

Here's the region for you:

Answer 14

although that should be not e^(-9/2), but e^(-9)