MHB 15.5.63 - Rewrite triple integral in spherical coordinates

[karush]

Write interated integrals in spherical coordinates
for the following region in the orders
$dp \, d\theta \, d\phi$
and
$d\theta \, dp \, d\phi$
Sketch the region of integration. Assume that $f$ is continuous on the region
\begin{align*}\displaystyle
I_{63}&=\int_{0}^{2\pi}\int_{\pi/6}^{\pi/2}\int_{2\csc{\phi}}^{4}
f(\rho,\phi,\theta)\rho^2
\,d\rho \,d\phi \,d\theta\\
\end{align*}
The standard equation of Sphere is:

\begin{align*}\displaystyle
f(\rho,\phi,\theta)
&=(\rho - \rho_0)^2 + (\phi - \phi_0)^2 + (\theta - \theta_0)^2 = \alpha ^2
\end{align*}

ok kinda ? what the $\rho^2$ would do to this

ok assume the new iterated sets would be according to d? set.

this was the pick for the graph I just picked bView attachment 7314

 

[HallsofIvy]

Re: 15.5.63 triple int of f(\rho,\phi,\theta)\ph^2

You appear to be confusing the equation of a sphere in Cartesian coordinates, x^2+ y^2+ z^2= r^2, with the equation of a sphere in spherical coordinates, \rho= r.

 

[karush]

Re: 15.5.63 triple int of f(\rho,\phi,\theta)\ph^2

\begin{align}\displaystyle
f_{12}&=(4s^2-as+5bs)+cs^2-d\\
&=40s^3+5s^2+s-20\\
&=-s^2-as+5bs+cs^2-40s^3-s-d=-20 \\
(s&=0) d=20\\
(s&=-1) -1+a-5b+c+40+1=0 \\
&a-5b+c=-40\\
(s&=1)-1+a+5b+c-40-1=0\\
&-a+5b+c =42\\
\therefore &c=2 \,b=8 \, a=1
\end{align}