Find the mass of a solid bounded by the cylinder r = sin θ, the planes z = 0, θ = 0, θ = π/3, and the cone z = r, if the density is given by p(r, θ, z) = 4r.

Question

Find the mass of a solid bounded by the cylinder [latex]r=\sin heta[/latex], the planes [latex]z=0, heta=0[/latex], [latex] heta=\pi / 3[/latex], and the cone [latex]z=r[/latex], if the density is given by [latex] ho(r, heta, z)=4 r[/latex].

Accepted Answer

We first note that the solid may be written as [latex]S=\left\{(r, heta, z): 0 \leq heta \leq \frac{\pi}{3}, 0 \leq r \leq \sin heta, 0 \leq z \leq r ight\}[/latex]

Then

[latex]\begin{aligned}\mu &=\int_{0}^{\pi / 3} \int_{0}^{\sin heta} \int_{0}^{r}(4 r) r d z d r d heta \&=\int_{0}^{\pi / 3} \int_{0}^{\sin heta}\left\{\left.4 r^{2} z ight|_{0} ^{r} ight\} d r d heta=\int_{0}^{\pi / 3} \int_{0}^{\sin heta} 4 r^{3} d r d heta \&=\int_{0}^{\pi / 3}\left\{\left.r^{4} ight|_{0} ^{\sin heta} ight\} d heta=\int_{0}^{\pi / 3} \sin ^{4} heta d heta=\int_{0}^{\pi / 3}\left(\frac{1-\cos 2 heta}{2} ight)^{2} d heta \&=\frac{1}{4} \int_{0}^{\pi / 3}\left(1-2 \cos 2 heta+\frac{1+\cos 4 heta}{2} ight) d heta \&=\left.\frac{1}{4}\left(\frac{3 heta}{2}-\sin 2 heta+\frac{\sin 4 heta}{8} ight) ight|_{0} ^{\pi / 3}=\frac{1}{4}\left(\frac{\pi}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{16} ight) \&=\frac{\pi}{8}-\frac{9 \sqrt{3}}{64} .\end{aligned}[/latex]

Question 5.7.1: Find the mass of a solid bounded by the cylinder r = sin θ, ...

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