Find the mass of the solid bounded by the paraboloid z = x^2 + y^2 and the plane z = 4 if the density at any point is proportional to the distance from the point to the z-axis.

Question

Find the mass of the solid bounded by the paraboloid [latex]z=x^{2}+y^{2}[/latex] and the plane [latex]z=4[/latex] if the density at any point is proportional to the distance from the point to the z-axis.

Accepted Answer

The solid is sketched in Figure 2. The density is given by [latex] ho(x, y, z)= \alpha \sqrt{x^{2}+y^{2}}[/latex], where [latex]\alpha[/latex] is a constant of proportionality. Thus since the solid may be written as [latex]S=\left\{(x, y, z):-2 \leq x \leq 2,-\sqrt{4-x^{2}} \leq y \leq \sqrt{4-x^{2}}, x^{2}+y^{2} \leq z \leq 4 ight\}[/latex]

we have

[latex]\mu=\iiint_{S} \alpha \sqrt{x^{2}+y^{2}} d V=\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \int_{x^{2}+y^{2}}^{4} \alpha \sqrt{x^{2}+y^{2}} d z d y d x[/latex]

We write this expression in cylindrical coordinates, using the fact that [latex]\alpha \sqrt{x^{2}+y^{2}}= \alpha r[/latex]. We note that the largest value of [latex]r[/latex] is 2, since at the "top" of the solid, [latex]r^{2}=x^{2}+y^{2}=4[/latex]. Then

[latex]\begin{aligned}\mu &=\int_{0}^{2 \pi} \int_{0}^{2} \int_{r^{2}}^{4}(\alpha r) r d z d r d heta=\int_{0}^{2 \pi} \int_{0}^{2} \alpha r^{2}\left(4-r^{2} ight) d r d heta \&=\alpha \int_{0}^{2 \pi}\left\{\left.\left(\frac{4 r^{3}}{3}-\frac{r^{5}}{5} ight) ight|_{0} ^{2} ight\} d heta=\frac{64 \alpha}{15} \int_{0}^{2 \pi} d heta=\frac{128}{15} \pi \alpha.\end{aligned}[/latex]

Question 5.7.2: Find the mass of the solid bounded by the paraboloid z = x^2...

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