Solution verification (volume)

Question

In the following example I am getting that volume of body bounded by surfaces $x^2+y^2=8, x^2+y^2+4x=0, z=10-x^2-y^2, x+2z=3$ is negative, so I want to know where am I mistaking. First those two cylinders will give me bounds for $x,y$. Since first one has center in $(0,0)$ and radius $\sqrt{8}$ and second center in $(-2,0)$, radius $2$. Now, after using cylindrical coordinates, first circle gives $r=2\sqrt{2}$ and the second $r=-4\cos\phi$. And then $-4\cos\phi\le r\le \sqrt{8}$ , $\frac{\pi}{2}\le \phi \le \frac{3\pi}{2}$(since $\cos\phi$ needs to be negative). And the I get volume $\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}d\phi\int\limits_{-4\cos\phi}^{\sqrt{8}} (10-r^2-\frac{3r\cos\phi}{2})rdr$

Answer 1

For the intersection of both cylinders, $-4 \cos \phi = 2 \sqrt 2 \implies \phi = - \frac{1}{\sqrt2}$ and you get $\phi = \frac{3 \pi}{4}, \frac{5 \pi}{4}$. Now for $\frac{3 \pi}{4} \leq \phi \leq \frac{5\pi}{4}, \,$ you are bound by $r = 2\sqrt2$ and for $\frac{\pi}{2} \leq \phi \leq \frac{3\pi}{4}, \, \frac{5\pi}{4} \leq \phi \leq \frac{3\pi}{2}$, $r = -4 \cos \phi$.
If you go by this, the bound of $z \,$ will be $\, 10-r^2 \leq z \leq \frac{3-r\cos \theta}{2}$ but this is not correct. For example, if you check for $r = \sqrt 8$ and $\theta = \pi$, you realize your lower and upper bounds are not correct.

So to find the right limits, let's find intersection points.

i) Intersection of plane and paraboloid for $x \lt 0$,

$z = 10 - x^2 - y^2 = \frac{3-x}{2} \implies y = \pm \sqrt {\frac{17+x}{2} - x^2}$. Also substituting $y = 0, $ we get $\frac{1 - \sqrt{137}}{4} \leq x$.

ii) Common intersection of paraboloid and cylinder with plane,

$y^2 = 10 - z - x^2 = -4x-x^2 \implies z = 4x + 10 = \frac{3-x}{2}, x = -\frac{17}{9}, $

So the integral to find volume $(V = V_1 + V_2)$ is as below -

$V_1 = \displaystyle 2 \, \int_{(1 - \sqrt{137}) / 4}^{-17/9} \int_0^{\sqrt{\frac{x+17}{2} - x^2}} \int_{(3-x)/2}^{10-x^2-y^2} \, dz \, dy \, dx$

$V_2 = \displaystyle 2 \, \int_{-17/9}^0 \int_0^{\sqrt{-4x-x^2}} \int_{(3-x)/2}^{10-x^2-y^2} \, dz \, dy \, dx$

$\,$

Some additional details if you require -

To visualize how the plane cuts these solids, refer to the above diagram. We need to keep the cylinder $x^2 + y^2 = 8$ aside for now. I will come to that later.

At any given $z,\,$ think of the circle centered at the origin as cross section of the paraboloid. If you start at $z = 10, \,$ the plane cuts the cylinder $x^2 + y^2 + 4x = 0$ outside the paraboloid. Then as $z$ reduces, the radius of the paraboloid increases and so does the area of intersection of the cylinder and paraboloid. At some point the plane cuts the paraboloid within the intersection that you see in the diagram. Then as you reduce $z$ further, the plane cuts the cylinder within the intersection that you see in the diagram.

Now coming to the cylinder $x^2+y^2 = 8,$ which is only relevant for $z \leq 2$ as that is when the radius of the paraboloid is $ \geq \sqrt 8$ but for $z \leq 2, $ the plane cuts the other cylinder to the right of the intersection of both cylinders and so if we find that bound, we will automatically be within the intersection of both cylinders.