Volume

Question

Volume

dramaqueen1201 · Answer

the answer is supposed to be 3pi/5 but i'm getting 4pi/5

i don't even think my method of doing it was good >.<

dramaqueen1201 · Answer

would the last example work similarly? http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx

ganeshie8 · Answer

May I see your work or can you describe your method briefly ?

ganeshie8 · Answer

like, have you used polar/cylindrical coordinates ?

dramaqueen1201 · Answer

I followed the last example from the link above. i used polar coordinates, yeah and in the example they already had some thing to integrate but since i didn't i just put that as one, but everything else, i did the same.

ganeshie8 · Answer

I'm getting 2pi/5 http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B2pi%7D%5Cint%5Climits_0%5E1+r%5E4+dr+d%5Ctheta

dramaqueen1201 · Answer

how did you get r^4?

ganeshie8 · Answer

Would you agree that the projection in xy plane is a unit disc ?

dramaqueen1201 · Answer

yes

ganeshie8 · Answer

Nvm, I've interpreted it worngly. Let's try agian.

ganeshie8 · Answer

Here the surface is f(x,y) = (x^2+y^2)^(3/2)

We're interested in the volume above the surface f(x,y) and below the plane z = 1

dramaqueen1201 · Answer

okay

ganeshie8 · Answer

Therefore the bounds for z are :

z = 1 - f(x,y)

ganeshie8 · Answer

$$V = \iint 1-f(x,y)\,dx\,dy$$

ganeshie8 · Answer

In polar coordinates : 

$$V = \int\limits_0^{2\pi}\int\limits_0^1 [1-(r^2)^{3/2}]~rdrd\theta$$

dramaqueen1201 · Answer

okay

ganeshie8 · Answer

Yaay that seems to give us 3pi/5 http://www.wolframalpha.com/input/?i=%5Cint%5Climits_0%5E%7B2pi%7D%5Cint%5Climits_0%5E1+(1-r%5E3)*r+dr+d%5Ctheta

dramaqueen1201 · Answer

ooooooh okay!!

dramaqueen1201 · Answer

Thank you so much! @ganeshie8

ganeshie8 · Answer

yw