Volumes help (1 Viewer)

-s- · Apr 18, 2017

Could someone please check if my answer is correct?

A tank with plane sides has square horizontal cross-sections whose sides cary in length from 2 m at the base to 5 m at the top; the height is 12 m. Find the total mass of liquid in the tank, if the density decreases uniformly with the height from 20kg/m^3 at the bottom to 12kg/m^3 at the top

$Let $ x(h)=ah+b $ be the side length of the square cross section at heigh $ h $. When h=0, x=2. When h=12, x=5 $\Rightarrow x=\frac{1}{4}h+2 \\ \\ \therefore \delta V= (\frac{1}{4}h+2)^2 \delta h \\ \Rightarrow V=\int_{0}^{12}\left ( \frac{1}{4}h+2 \right )^2 $ d$h = 156$ m$^3 \\ \\ \frac{\mathrm{d} \rho}{\mathrm{d} h} = -k \Rightarrow \rho = -hk+c$ When $h=0, \rho=20. $ When $h=12, \rho =12 \\ \Rightarrow \rho=-\frac{2}{3}h+20. \\ \\\therefore $ Density of the whole solid $=\int_{0}^{12}\left ( -\frac{2}{3}h+20 \right )$ d$h = 192$ kgm$^{-3} \\ \\ \rho = \frac{m}{v} \Rightarrow m = \rho v = 192\cdot 156 = 29952 $ kg$

InteGrand · Apr 18, 2017

-s- said:
Could someone please check if my answer is correct?

A tank with plane sides has square horizontal cross-sections whose sides cary in length from 2 m at the base to 5 m at the top; the height is 12 m. Find the total mass of liquid in the tank, if the density decreases uniformly with the height from 20kg/m^3 at the bottom to 12kg/m^3 at the top

$Let $ x(h)=ah+b $ be the side length of the square cross section at heigh $ h $. When h=0, x=2. When h=12, x=5 $\Rightarrow x=\frac{1}{4}h+2 \\ \\ \therefore \delta V = (\frac{1}{4}h+2)^2 \delta h \\ \Rightarrow V=\int_{0}^{12}\left ( \frac{1}{4}h+2 \right )^2 $ d$h = 156$ m$^3 \\ \\ \frac{\mathrm{d} \rho}{\mathrm{d} h} = -k \Rightarrow \rho = -hk+c$ When $h=0, \rho=20. $ When $h=12, \rho =12 \\ \Rightarrow \rho=-\frac{2}{3}h+20. \\ \\\therefore $ Density of the whole solid $=\int_{0}^{12}\left ( -\frac{2}{3}h+20 \right )$ d$h = 192$ kgm$^{-3} \\ \\ \rho = \frac{m}{v} \Rightarrow m = \rho v = 192\cdot 156 = 29952 $ kg$

$\noindent The approach should be essentially as follows. Consider slicing the solid into thin slices parallel to the base. We know that a small slice between height $h$ and $h+\delta h$ has volume $\delta V \approx x^{2}(h)\delta h$ (where $x^{2}(h)\equiv (x(h))^{2}$), so the mass of solid in this slice is approximately $\delta M \approx \rho (h)\delta V \approx \rho (h)x^{2}(h)\delta h$. (This is because the density throughout this small slice is approximately constant at $\rho (h)$.) The total mass is then found by summing masses from all slices and letting $\delta h \to 0$, which means we end up integrating over all possible values of $h$ to get a total mass of $M = \int_{0}^{12} \rho(h)x^{2}(h)\,\mathrm{d}h$. If you wanted to use your approach, you'd need to find the ``average density'' to use as your $\rho$ in $M = \rho V$.$

-s- · Apr 18, 2017

InteGrand said:
$\noindent The approach should be essentially as follows. Consider slicing the solid into thin slices parallel to the base. We know that a small slice between height $h$ and $h+\delta h$ has volume $\delta V = x^{2}(h)\delta h$ (where $x^{2}(h)\equiv (x(h))^{2}$), so the mass of solid in this slice is approximately $\delta M \approx \rho (h)\delta V = \rho (h)x^{2}(h)\delta h$. (This is because the density throughout this small slice is approximately constant at $\rho (h)$.) The total mass is then found by summing masses from all slices and letting $\delta h \to 0$, which means we end up integrating over all possible values of $h$ to get a total mass of $M = \int_{0}^{12} \rho(h)x^{2}(h)\,\mathrm{d}h$. If you wanted to use your approach, you'd need to find the ``average density'' to use as your $\rho$ in $M = \rho V$.$

oh yes thank you for that

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