Question

triple integral

Answer 1

So the parabaloid is symmetric around the z axis and is opening downwards. So this implies:
\[0 \le \phi \le 2 \pi; 0 \le z \le 9-r^2;0 \le r \le 3\]
So we have:
\[\int\limits_0^{2 \pi} \int\limits_0^3 \int\limits_0^{9 - r^2} r^{1/2} r dz dr d \phi\]
Where r dz dr dphi is the dV for cylindrical.

Answer 2

And the r bounds come from setting 0=9-r^2 implies r=+-3 but r is always positive.

Answer 3

@malevolence19 i got(18(3^2.5-1)/5-2/9(3^4.5-1))2pi as the final answer, its really strange

Answer 4

Well phi is independent so you have:
\[2 \pi \int\limits_0^3 \int\limits_0^{9 - r^2}r^{3/2}dz dr d \theta=2 \pi \int\limits_0^3 r^{3/2}(9-r^2)dr=2 \pi (\frac{18 r^{5/2}}{5}-\frac{2 r^{9/2}}{9})_0^3\]

Answer 5

@malevolence19 our answers r same

Answer 6

0^n=1 right?

Answer 7

0^n is zero. n^0 is 1 for n not equal to 0.

Answer 8

sry!!!! ireally made a very stupid midtake! i just 2 sleepy. thx a lot !!!!