Multiple integration questions and my attempt P1: Find volume of portion of sphere centered at $(0,a,0)$ with radius $a$,  between planes $y=0$ and $y=a$ .
I changed to spherical coordinates and I calculated volume in 1st quadrant and multiplied by 4 to  get required volume. The limits for $\theta$ and $\phi$ are $0$ to $\pi/2$ and for $\rho$ are $0$ to $2\sin\theta \sin \phi$, but I couldn't get correct and and is double than what it is. Where am I wrong ? Can please someone help me figure out .
P2: 
Find volume of portion of sphere which is inside right circular cone having its vertex at origin and making angle alpha with positive z axis ..
in this q i dont know bounds..can someone help me with bounds only and explanation of them
P3: Find volume of solid which is below plane z=2x+3 and in xy plane bounded by $y^{2}=x$ , $x=0$ , $x=2$ ..i cant figure out region to integrate over in xy plane
 A: For part 1, you have correctly determined that the spherical coordinates of
a point on the sphere are $2a \sin \phi \sin \theta.$
(You don't write the factor of $a$ but I assume you are scaling the region down
by that factor and that you scale up your result appropriately.)
The problem is that you are now computing the entire volume of the sphere,
whereas you are only supposed to find the volume of the hemisphere between
the planes $y=0$ and $y=a.$
For chords of the sphere that pass through the disk $x^2 + z^2 \leq a^2$ in the
plane $y = a,$ you are integrating over the whole length of the chord whereas
you really should integrate only over the part for which $y \leq a.$
If you're going to do this with spherical coordinates around the $z$-axis,
it means splitting your region of integration in two parts in a somewhat
complicated way.
For $\theta$ such that $cos^2\theta \geq \frac{1-2\cos^2\phi}{2\sin^2\phi}$
(if I've done the calculation correctly--you should check this), you can integrate
over the sphere as you are doing. But for other $\theta$ you want to be integrating
over a right circular cone whose base is in the plane $y=a.$
For that region, the range of $\rho$ is from $\rho = 0$ to
$\rho = \frac{a}{\sin\phi\sin\theta}.$
This is much easier if you take spherical coordinates around the $y$-axis
instead of the $z$-axis.
If $\phi$ is measured from the $y$-axis then $\theta$ no longer appears in
the equation of the sphere; $\rho$ is bounded by the plane $y=a$
whenever $0 \leq \phi \leq \frac\pi4,$ and $\rho$ is bounded by the surface
of the sphere for $\frac\pi4 \leq \phi \leq \frac\pi2.$
Presumably there is a point to this exercise
other than just getting the answer (since the answer is so easy to get without
doing any integration at all!), so you'll have to decide how to do it,
but if "spherical coordinates" were specified without specifying the axis,
perhaps this is an acceptable simplification.  (Perhaps the point is to understand
the difference it makes to choose an appropriate axis for the spherical coordinates,
in which case the $y$-axis is probably the intended choice!)
For part 2, the cone consists simply of all points for which $\phi=\alpha.$
The interior of the cone is all points where $\phi<\alpha.$
(In this case $\phi$ is measured from the $z$-axis; anything else would make this
integration much harder, I think.)
It should be fairly obvious how this affects the region of integration.
The range of $\rho$ is still $0$ to $2a \sin \phi \sin \theta,$ as you previously
determined, and you can still take $\theta$ from $0$ to $\frac\pi2.$
I think the main difference is you take $\phi$ only from $0$ to $\alpha$;
also, unless this was meant to be a double-ended cone, this integral would
cover half the volume (which is all above the $x,y$ plane along with the cone),
so you would only multiply by two rather than four.
If the cone is double-ended, however, you would still have the four-way symmetry
and would still multiply by four.
For part 3, the plot of $y^2 = x$ in the $x,y$ plane is as shown in the
figure below 
(courtesy of https://www.desmos.com/calculator, a very handy tool to remember).
The lines $x=0$ and $x=2$ are also shown; the region bounded by all these
curves should be clear enough.

