Question

Consider the solid that is bounded below by the cone \(z = \sqrt{3x^{2}+3y^{2}}\)
and above by the sphere \(x^{2} +y^{2} + z^{2} = 16.\).Set up only the appropriate triple integrals in cylindrical and spherical coordinates needed to find the volume of the solid.

Answer 1

Step 1
To set the triple integral in cylindrical coordinates
By using relation,
\(x = r \cos \theta\)
\(y = r \sin \theta\)
\(z = z\)
Thus,
The cone \(z = \sqrt{3x^{2}+3y^{2}}\) in cylindrical coordinate becomes,
\(z = \sqrt{3r^{2}}=\sqrt{3r}\)
And the sphere become \(r^{2} + z^{2} = 16\)
To find the limit of r,
Consider,
\(\Rightarrow\sqrt{3r^{2}}=\sqrt{16-r^{2}}\)
\(\Rightarrow 3r^{2} = 16 - r^{2}\)

\(\Rightarrow r^{2} = 4\)
\(\Rightarrow r = 2\)
Step 2
Thus, we can describe the region as
\(E = \{(r, \theta, z)|0 \leq \theta \leq 2 \pi|0 \leq r \leq 2 \sqrt{3r} \leq z \leq \sqrt{16-r^{2}}\}\)
Hence, the triple integral for the volume by cylindrical coordinates is
\(V=\int_{0}^{2 \pi} \int_{0}^{2} \int_{\sqrt{3r}}^{\sqrt{16-r^{2}}} r d z d r d \theta\)
Step 3
Now, to set the triple integral in spherical coordinates
Since,
\(p^{2}=x^{2}+y^{2}+z^{2}\)
\(\tan \theta = \frac{y}{x}\)
\(\varphi= \arccos (\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}})\)
The sphere \(x^2 + y^2 + z^2 = 16\ gives\ p^2 = 16 => p = 4\)
From the cone
\(z=\sqrt{3x^{2}+3y^{2}}=\sqrt{3r}\)
\(\Rightarrow p \cos (\varphi)=\sqrt{3}p \sin (\varphi)\)
\(\Rightarrow p \tan (\varphi)=\frac{1}{\sqrt{3}}\)
\(\varphi = \frac{\pi}{6}\)
Step 4
Thus, we can describe the region as
\(E = \{(p, \varphi, \theta)|0 \leq p \leq 4, 0 \leq \varphi \leq \frac{\pi}{6}, 0 \leq \theta \leq 2 \pi\}\)
Hence, the triple integral for the volume of the solid by spherical coordinate is
\(V=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{6}} \int_{0}^{4} p^{2} \sin (\varphi)\ d\ pd\ \varphi\ d\ \theta\)
Step 5
Now, evaluating the integral of cylindrical coordinate we get.
\(\Rightarrow V = \int_{0}^{2 \pi} \int_{0}^{2} \int_{\sqrt{3 r}}^{\sqrt{16-r^{2}}} rdzdrd \theta\)
\(\Rightarrow V = 17.9582\)
And evaluating the integral of spherical coordinate
\(\Rightarrow V = \int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{6}} \int_{0}^{\sqrt{4}} r^{2} \sin (\varphi) dpd \varphi d \theta\)
\(\Rightarrow V = 17.9582\)
Thus, by both coordinate systems, we get the same volume.
Therefore, both triple integrals are appropriate.