Volume of $ \Gamma=\{(x,y,z)\in \mathbb{R}^3:\sqrt{x^2+y^2}\le z\le \sqrt{2x^2+2y^2};x^2+y^2+z^2\le 3\} $

Question

I evaluated the volume of the set $$ \Gamma=\{(x,y,z)\in \mathbb{R}^3:\sqrt{x^2+y^2}\le z\le \sqrt{2x^2+2y^2};x^2+y^2+z^2\le 3\} $$ by using the Pappu's centroid theorem, but I'm in trouble while evaluating the volume without using this theorem... can anybody help me? Thank you!

Answer 1

Here are some helps. You say you know cylindrical coordinates, so I'll use that approach (though there are several other approaches such as washers, cylindrical shells, spherical coordinates, and Pappus's theorem).

As you can see from this side view (the eye on the plane $z=0$), the desired volume is between two cones and inside a sphere. We use cylindrical coordinates, and rewrite your inequalities using $r=\sqrt{x^2+y^2}$.

$$r \le z \tag{1}$$ $$z \le \sqrt 2r \tag{2}$$ $$r^2+z^2 \le 3 \tag{3}$$

Inequality (1) implies that $z$ is nonnegative, so we need look only at the upper half of the diagram. We can rewrite those inequalities to put bounds on $r$:

$$r \ge \frac{z}{\sqrt 2} \qquad\text{from (2)} \tag{4}$$ $$r \le z \qquad\text{copying (1)} \tag{1}$$ $$r \le \sqrt{3-z^2} \qquad\text{from (3)} \tag{5}$$

The diagram shows that there are two regions. In the first region, the "wedge," $r$ is limited by the cones defined by inequalities $(4)$ and $(1)$. That region begins at the bottom at $z=0$ and stops at the top where the cone in $(1)$ intersects the sphere in $5$. We can find that value of $z$ be equating the two expressions:

$$z=\sqrt{3-z^2}$$

The solution to that is $z=\sqrt{\dfrac 32}$. So for $0\le z\le \sqrt{\dfrac 32}$ we have $\dfrac{z}{\sqrt 2}\le r\le z$. You can integrate with those limits on $z$ and $r$, and $0\le \theta\le 2\pi$. This gives us the integral

$$V_1=\int_{0}^{\sqrt{3/2}} \int_{z/\sqrt 2}^z \int_{0}^{2\pi} r\,d\theta\,dr\,dz$$

There is another region above that, the "ring" between the inner cone and the sphere. The bottom of that region is again $z=\sqrt{\dfrac 32}$. The top is the intersection of the cone defined by $(4)$ and the sphere defined by $(5)$. We find that upper value of $z$ again by equating expressions:

$$\frac{z}{\sqrt 2}=\sqrt{3-z^2}$$

The solution to that is $z=\sqrt 2$. So for $\sqrt{\dfrac 32}\le z\le \sqrt 2$ we have $\dfrac{z}{\sqrt 2}\le r\le \sqrt{3-z^2}$. Again, you can integrate with those limits on $z$ and $r$, and $0\le \theta\le 2\pi$. This gives us the integral

$$V_2=\int_{\sqrt{3/2}}^{\sqrt 2} \int_{z/\sqrt 2}^{\sqrt{3-z^2}} \int_{0}^{2\pi} r\,d\theta\,dr\,dz$$

Find those two volumes, add them, and you have your volume.

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In a comment you asked if the same technique would work if both square roots were removed from the limiting inequalities. The answer is yes, but the limits on the integrals would change.

The resulting limits on $r$ would then be

$$r\le\sqrt z, \quad, r\ge\sqrt{\frac z2}, \quad, r\le\sqrt{3-z^20}$$

We again get two regions. The lower limit of the first is $z=0$. The upper limit is given by the combination of the first and last inequalities, which gives the upper limit

$$z=\frac{-1+\sqrt{13}}{2}$$

The second region's lower limit is the first's upper limit. The second's upper limit is given by the last two inequalities, which gives $z=\frac 32$. So your two integrals would be

$$V_1=\int_0^{(-1+\sqrt{13})/2}\int_{\sqrt{z/2}}^{\sqrt z}\int_0^{2\pi} r\,d\theta\,dr\,dz$$

$$V_2=\int_{(-1+\sqrt{13})/2}^{3/2}\int_{\sqrt{z/2}}^{\sqrt{3-z^2}}\int_0^{2\pi} r\,d\theta\,dr\,dz$$

Answer 2

In case it's of interest, this integral can be done in spherical coordinates. If $\theta$ denotes longitude and $\phi$ denotes colatitude, the solid $\Gamma$ is defined by $$ 0 \leq \theta \leq 2\pi,\quad \arctan(1/\sqrt{2}) \leq \phi \leq \pi/4,\quad 0 \leq \rho \leq 3; $$ the volume of $\Gamma$ is therefore $$ \int_{0}^{2\pi} \int_{\arctan(1/\sqrt{2})}^{\pi/4} \int_{0}^{3} \rho^{2} \sin\phi\, d\rho\, d\phi\, d\theta, $$ or

$18\pi\left(\dfrac{1}{\sqrt{2}} - \dfrac{\sqrt{2}}{\sqrt{3}}\right)$.