Volume of body bounded with surfaces

Question

I need to find volume bounded with $x=2y-y^2-z^2$ and $2y-x-1=0$. So first one is paraboloid that is located on x-axis and its "maximum" is in $(0,2,0)$ and the second is plane paralele to z-axis. I just want to check do we get the volume when substract volume of $2y-x-1=0$ from paraboloid.(Also using polar coordinates it is easy to get $0\le r\le 1, 0\le \phi \le 2\pi$). I'm just not sure whether I sketched this correctly, precisely is given plane above paraboloid?

Answer 1

You can set up the integral in different orders. A few of the examples are given below. But for some clarity on the volume bound between the plane and the paraboloid, they intersect only between $-3 \leq x \leq 1$. The two extreme points of intersection are $(1, 1, 0)$ and $(-3, -1, 0)$. In between, as you rightly pointed out, the intersection points are on $y^2 + z^2 = 1$

Integral setup -

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} \int_{2y-1}^{2y-y^2-z^2} dx \, dz \, dy$

or

$\displaystyle \int_{-1}^{1} \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} \int_{2y-1}^{2y-y^2-z^2} dx \, dy \, dz$

In cylindrical coordinates -

$\displaystyle \int_{0}^{2\pi} \int_{0}^{1} \int_{2r\sin \theta - 1}^{2r \sin \theta - r^2} r \, dx \, dr \, d\theta$

Answer 2

Sometimes viewing the volume helps:

Answer 3

The intersection between $x = 2 y - 1$ and $x=-y^2+2 y-z^2$ is the circle

$y^2+z^2=1$, which means that the limits for $z$ and $y$ are $$-\sqrt{1-y^2}\le z\le \sqrt{1-y^2};\;-1\le y\le 1$$ Thus the volume is $$V=\int _{-1}^1\int _{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int _{2 y-1}^{-y^2+2 y-z^2}dxdzdy=\frac{\pi}{2}$$

Hope this helps

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