1. Help determining volume

I have the equation ...

Determine the volume for $\displaystyle f(x,y)=x^2+y^2+1$; 0<x<3 (greater than or equal to on the <, I didn't know how to do that); 0<y<3 (also greater than or equal to)

2. Hello,

Wouldn't it be $\displaystyle \int_0^{2 \pi} \int_0^3 \int_0^3 x^2+y^2+1 \ dx dy dz$ ?

3. Originally Posted by Moo
Hello,

Wouldn't it be $\displaystyle \int_0^{2 \pi} \int_0^3 \int_0^3 x^2+y^2+1 \ dx dy dz$ ?
Yeah, that looks about right. Thanks, it's a cylinder it seems. Thanks for help with the setup I always stink at remembering form formulas

4. Be careful of that z limit. It wouldn't be 2Pi, unless you're using polar.

$\displaystyle z=x^{2}+y^{2}+1$

Perhaps try this:

$\displaystyle \int_{0}^{3}\int_{0}^{3}\int_{0}^{x^{2}+y^{2}+1}dz dydx$

A double integral may be better:

$\displaystyle \int_{0}^{3}\int_{0}^{3}[x^{2}+y^{2}+1]dydx$

5. I'm not used to it, can a volume be obtained with a double integral ?

6. Fornuately, for me the professor gave us a solution set

a.) 62
b.) 63
c.) 64
d.) None of these

so if I'm in the ballpark, I'll know it. I might just try both

7. Originally Posted by Moo
I'm not used to it, can a volume be obtained with a double integral ?
Well if you look at the change the only difference is he evaluated the innermost integral leaving you with two instead of three

8. Yes, double integrals give volume. Look at choice b

9. Oh yeah indeed... I really have to learn my lessons
Thanks !