Question

can someone help me with some spherical coordinates?

Answer 1

i have this integral \[\int\limits \int\limits \int\limits dx dy dz\]

Answer 2

\[V: x^2+y^2+z^2\le 2x\]

Answer 3

@Loser66 Do you know any of this?

Answer 4

for the dx one i think it's [0,2] for dy [0,2pi] but for the third one i can't figure it out not even if it hits me

Answer 5

Are you sure that the right side of the inequality is 2x ?

Answer 6

sorry it;s 2z

Answer 7

This implies:

\( \large x^2+y^2 + z^2 \le 2z\) and so \( \large z^2-2z<-(x^2+y^2)\) this implies that \(\large z^2 < 2z \) or \( z<2\).

Therefore, \( \large x^2+y^2 + z^2 \le 2z\le4\)

Using spherical coordinates, we have:

$$
\large{
x=4\sin \phi \cos \theta\\
y=4\sin \phi \sin \theta\\
z=4\cos\phi\\
dV=4^2\sin\phi ~d\rho~d\phi~d\theta
}
$$
With limits:
$$
\large{
0 \le \rho \le 4\\
0 \le \phi \le \pi\\
0 \le \theta\le 2\pi\\
}
$$

Answer 8

Correction:

$$
\large{
x=\rho\sin \phi \cos \theta\\
y=\rho\sin \phi \sin \theta\\
z=\rho\cos\phi\\
dV=\rho^2\sin\phi ~d\rho~d\phi~d\theta
}
$$

Answer 9

thank you :D

Answer 10

yw