Find the volume enclosed below the plane $3x+4y+z=12$ bounded by the region between $x^2+y^2=2x$ and above the $xy$ plane.
Attempt: I need a little help deciding the limits of integration w.r.t $\theta$ when $x,y$ are changed to polar coordinates. So, here's how I proceeded:
$x^2+y^2=2x ~~~ \equiv ~~~ (x-1)^2+y^2=1 ~~~\equiv~~~r=2 \cos\theta$
which is a circle with centre $(1,0)$ and radius $1$. The $y$ axis is a tangent to this circle.
Thus, $V = \int \int \int dv = \int \int_D \int_{0}^{12-3x-4y} dz ~dx ~dy $
Where $D$ refers to the projection of the solid of interest on the $xy$ plane.
$V=\int \int_D (12-3x-4y)dx~dy$
Changing the above integral to polar coordinates :
$V = \int_{-\pi/2}^{\pi/2}~~\int_0^{2 \cos \theta}~~(12-3r \cos \theta - 4 r \sin \theta)~r dr~d\theta$. But, my textbook says the limits of $\theta$ to lie between $0$ and $\pi$. I feel that any angle beyond $\dfrac {\pi}{2}$ takes us beyond the region $D$ into the second quadrant.
Could someone please clarify why $\theta$ from $-\pi/2$ to $\pi/2$ might be incorrect and $0$ to $\pi$ might be the correct one !. Thanks a lot!
Recenter the cylinder with the variable change $u=x-1$. Then, its equation simplifies to $u^2+y^2=1$ and the capping plane becomes $3u+4y+z=9$.
Now, integrate the volume in the cylindrical coordinates as follows,
$$V= \int_{u^2+y^2\le1} (9-3u-4y)dudy=\int_0^{2\pi}\int_0^1(9-3r\cos\theta-4r\sin\theta)rdr\theta=9\pi$$