mass of a body with a density function

Question

Let $V$ be the body confined by the following quadratic surfaces:
$$ x^{2} + y^{2} = 12z,\quad x^{2} = 3z,\quad z = 3 $$

• Compute the mass of $V$ with the density function $\rho = 1$.
• My problem is I can't figure the integration boundaries$\ldots$
• I know that substituting to spherical or cylindrical coordinates won't help because it says so in the question.

Answer 1

lower- and upper-bounds for the mass of the body

$M \in (M_{l},M_{u})=(93.53,108)$.

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numerical verification

From numerical calculations, I obtained $M \approx 101.17 \in \left( 100.77, 108 \right)$, the improved lower- and upper-bounds mentioned in comments.

Plot of the surface the region:

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calculation

Starting from $x^2 + y^2 = 12z$ and $x^2 = 3z$, we obtain $y^2 \in (9z,12z)$. This simplifies the problem a bit. To visualize the problem, take different values of $z$ and solve for $x$ and $y$,

$$ \begin{align} & z=1 \implies |x| < \sqrt{3}, 3<|y|<2\sqrt{3} \\ & z=2 \implies |x| < \sqrt{6},3\sqrt{2}<|y|<2\sqrt{6} \\ & z=3 \implies |x| < 3, 3\sqrt{3}<|y|<6 \end{align} $$

calculating a lower-bound for the mass of the body

Going along the $z$-axis, the area of the largest rectangle inside the truncated circle (for obtaining a lower-bound to the area $A_{l}$) is

$$ \begin{align} A_{l}(z) &= \left( 2\times 3\sqrt{z} \right) \times \left( 2\times \sqrt{3z} \right) \\ &= 12\sqrt{3}z \end{align} $$

What remains is to add the height element to this rectangle which is $dz$:

$$ \begin{align} dV_{l} &= A_{l}(z)dz \\ &= 12\sqrt{3}zdz \end{align} $$

Since mass is volume times density, which is just $1$ here, numerically, $dM_{l}=dV_{l}$.

Integrate along $z$ from $0$ (since $z$ is a sum of non-negative quantities, it is also non-negative) to $3$ to obtain the volume:

$$ \begin{align} M_{l} &= \int_{z=0}^{3} dM_{l} = \int_{z=0}^{3} dV_{l} = \int_{z=0}^{3} A_{l}(z)dz = \int_{z=0}^{3}12\sqrt{3}zdz \\ &= 12\sqrt{3}\int_{z=0}^{3}zdz \\ &= 12\sqrt{3}\left[ \dfrac{9}{2} \right] \\ &= 54\sqrt{3} \approx 93.53 \end{align} $$

calculating an upper-bound for the mass of the body

Having found a lower-bound to the mass, let's find an upper-bound. Along the $y$-axis, the larger smallest rectangle containing the horizontal sections has boundaries $y = \pm 2\sqrt{3z}$. We may calculate an upper-bound to the area as well.

Going along the $z$-axis, the area of the smallest rectangle containing the truncated circle (for obtaining an upper-bound to the area $A_{u}$) is

$$ \begin{align} A_{u}(z) &= \left( 4\times \sqrt{3z} \right) \times \left( 2\times \sqrt{3z} \right) \\ &= 24z \end{align} $$

$$ \begin{align} dV_{u} &= A_{u}(z)dz \\ &= 24zdz \end{align} $$ $$ dM_{u}=dV_{u} $$ $$ \begin{align} M_{u} &= \int_{z=0}^{3} dM_{u} = \int_{z=0}^{3} dV_{u} = \int_{z=0}^{3} A_{u}(z)dz = \int_{z=0}^{3}24zdz \\ &= 24\int_{z=0}^{3}zdz \\ &= 24 \left[ \dfrac{9}{2} \right] \\ &= 108 \end{align} $$

$M \in (M_l,M_u)=(54\sqrt{3},108)$. That's not bad.

Answer 2

$x^2+y^2 = 12$ describes a paraboloid

$x^2 = 3z$ is a parabolic cylinder
$z = 3$ is a plane.

How do the 3 surfaces intersect? Here is my picture.

plug $z = 3$ into the other two surfaces to get
$x^2+y^2 = 36, z=3$
$x = \pm 3, z=3$

And $x^2 = 3z$ into the equation of the paraboloid. $3z+y^2 = 12z$
$y^2 = 9z$

I would use this integral.

$M = \rho V = 4\int_0^3 \int_0^{3\sqrt z} \int_{\sqrt{3z}}^{\sqrt {12z-y^2}} \ dx\ dy\ dz$

or

$4\int_0^3 \int_{\sqrt{3z}}^{\sqrt{12z}} \int_{0}^{\sqrt {12z-x^2}} \ dy\ dx\ dz$

I am getting $108( \frac {\pi}{3} - \frac{\sqrt{3}}{4})$

In cylindrical coordinates, I have

$4\int_0^3\int_{0}^{\frac {\pi}{3}}\int_{\sqrt{3z}\sec{\theta}}^{\sqrt{12 z}} r\ dr\ d\theta\ dz$

Answer 3

The formula for volume computations with $xy$-projection is $$V=\sum_{i=1}^n\iint_{R_i}({z_{i,max}}-{z_{i,min}})dA.$$ In our example, we have three surfaces: $z=3,z=\frac{r^2}{12}$ and $z=\frac{r^2\cos^2\theta}{3}$ in cylindrical coordinates and $dA=rdrd\theta$.

Let me determine $R_i$s if I can, as I can only imagine the solid: I can not sketch the solid but I can work algebraically. First of all, we find the intersection of surfaces: $1)$ $z=3$ and $z=\frac{r^2}{12}$ gives $r=6$. $2)$ $z=\frac{r^2}{12}$ and $z=\frac{r^2\cos^2\theta}{3}$ gives $\theta=\frac{\pi}{3}$ $3)$ $z=3$ and $z=\frac{r^2\cos^2\theta}{3}$ gives $r\cos\theta=3$, in $[0,\frac{\pi}{2}]$.

Hence, I think, a correct subdivision of quarter of the solid is the following:

$1)$ $R_1: 0\leq\theta\leq\frac{\pi}{3}\;,0\leq r\leq \frac{3}{\cos\theta}\;$, $\frac{r^2\cos^2\theta}{3}\leq z\leq 3$.

$2)$ $R_2: \frac{\pi}{3}\leq\theta\leq\frac{\pi}{2}\;,0\leq r\leq 6\;$, $\frac{r^2}{12}\leq z\leq 3$.

Therefore, due to symmetry w.r.t. $x$ and $y$ axes, $$V=4\left(\int_{0}^{\frac{\pi}{3}}\int_0^{\frac{3}{\cos\theta}}(3-\frac{r^2\cos^2\theta}{3})rdrd\theta+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\int_0^6(3-\frac{r^2}{12})rdrd\theta\right)=27\sqrt{3}+18\pi.$$