[latex]{\displaystyle\int^{{2}{\pi}}_{0}} \ {\displaystyle\int^{{\pi}/2}_{0}} \ {\displaystyle\int^{1}_{0}} \ {\rho}^{2}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}} = {\displaystyle\int^{{2}{\pi}}_{0}} \ {{d}{\theta}} \ {\displaystyle\int^{{\pi}/2}_{0}} \ {\sin} \ {\varphi} \ {{d}{\varphi}} \ {\displaystyle\int^{1}_{0}} \ {\rho}^{2}{d}{\rho} = {({2}{\pi})}{(1)}{\left ( {\frac{1}{3}} \right )} = {\frac{{2}{\pi}}{3}}.[/latex]

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: [latex]{z} = {\sqrt{{3}{({x^2}+{y^2})}}}[/latex] or [latex]{\rho} \ {\cos} \ {\varphi} = {\sqrt{3}} \ {\rho} \ {\sin} \ {\varphi}[/latex] or [latex]{\tan} \ {\varphi} = {\frac{1}{\sqrt{3}}}[/latex] or [latex]{\varphi} = {\frac{\pi}{6}}[/latex].

For the sphere: [latex]{z} = {\sqrt{{4}-{x^2}-{y^2}}}[/latex] or [latex]{z^2} + {x^2} + {y^2} = {4}[/latex] or [latex]{{\rho}^{2}} = {4}[/latex] or [latex]{\rho} = {2}[/latex].

Thus, the triple integral for the volume is [latex]{{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ \ {\displaystyle\int^{{\varphi} = {{\pi}/{6}}}_{{\phi} = {0}}} \ \ {\displaystyle\int^{{\rho} = {2}}_{{\rho} = {0}}} \ {{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}[/latex].

1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.
   For the sphere:

   [latex]\begin{aligned} {x^2} + {y^2} + \ & {z^2} = {z} \\ & {{\rho}^{2}} = {\rho} \ {\cos} \ {\varphi} \\ & {\rho} = {\cos} \ {\varphi}. \end{aligned}[/latex]

   For the cone:

   [latex]\begin{aligned} {z} & = {\sqrt{{x^2}+{y^2}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\sqrt{{{\rho}^{2}}{{\sin}^{2}}{\varphi}{{\cos}^{2}}{\phi}+{{\rho}^{2}}{{\sin}^{2}}{\varphi}{{\sin}^{2}}{\phi}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\sqrt{{{\rho}^{2}}{{\sin}^{2}}{\varphi}{({{\cos}^{2}}{\phi}+{{\sin}^{2}}{\phi})}}} \\ {\rho} \ {\cos} \ {\varphi} & = {\rho} \ {\sin} \ {\varphi} \\ {\cos} \ {\varphi} & = {\sin} \ {\varphi} \\ {\varphi} & = {{\pi}/{4}}. \end{aligned}[/latex]

   Hence the integral for the volume of the solid region [latex]E[/latex] becomes

   [latex]{{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\varphi} = {{\pi}/{4}}}_{{\varphi} = {0}}} \ {\displaystyle\int^{{\rho} = {{\cos}{\varphi}}}_{{\rho} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\rho}} \ {{d}{\varphi}} \ {{d}{\theta}}}.[/latex]

2. Consider the [latex]{\varphi}{\rho}[/latex]-plane. Note that the ranges for [latex]{\varphi}[/latex] and [latex]{\rho}[/latex] (from part a.) are

   [latex]\begin{aligned} {0} & \ {\leq} \ {\varphi} \ {\leq} \ {{\pi}/4} \\ {0} & \ {\leq} \ {\phi} \ {\leq} \ {{\cos}{\varphi}}. \end{aligned}[/latex]

   The curve [latex]{\rho} = {\cos}{\varphi}[/latex] meets the line [latex]{\varphi} = {{\pi}/{4}}[/latex] at the point [latex]({{\pi}/{4}},{{\sqrt{2}}/{2}})[/latex]. Thus, to change the order of integration, we need to use two pieces:

   [latex]\hspace{8cm}\begin{align}&0\leq\rho\leq\sqrt2/2 \\&0\leq\varphi\leq\pi/4\end{align}[/latex] [latex]\text{ and }[/latex] [latex]\begin{align}&\sqrt2/2\leq\rho\leq1 \\&0\leq\varphi\leq\cos^{-1}\rho\end{align}[/latex]

   Hence the integral for the volume of the solid region [latex]E[/latex] becomes

   [latex]{{V}{(E)}} = {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\rho} = {{\sqrt{2}/{2}}}}_{{\rho} = {0}}} \ {\displaystyle\int^{{\varphi} = {{\pi}/{4}}}_{{\varphi} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}} \ + {\displaystyle\int^{{\theta} = {{2}{\pi}}}_{{\theta} = {0}}} \ {\displaystyle\int^{{\rho} = {1}}_{{\rho} = {{\sqrt{2}/{2}}}}} \ {\displaystyle\int^{{\varphi} = {{\cos}^{-1}}{\rho}}_{{\varphi} = {0}}} \ {{{\rho}^{2}}{\sin} \ {\varphi} \ {{d}{\varphi}} \ {{d}{\rho}} \ {{d}{\theta}}}.[/latex]

   In each case, the integration results in [latex]{{V}{(E)}} = {\frac{\pi}{8}}[/latex].

The ranges of the variables are

[latex]\begin{aligned} {-1} & \ \ \ {\leq} \ \ \ {y}{\leq}{1} \\ {0} & \ \ \ {\leq} \ \ \ {x}{\leq}{\sqrt{{1} - {{y}^{2}}}} \\ {{x}^{2}} + {{y}^{2}} & \ \ \ {\leq} \ \ \ {z}{\leq}{\sqrt{{{x}^{2}}+{{y}^{2}}}}. \end{aligned}[/latex].

The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex]{\theta}[/latex] and [latex]r[/latex] are

[latex]{-{\frac{\pi}{2}}} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{2}} \ {\text{and}} \ {0} \ {\leq} \ {r} \ {\leq} \ {1}.[/latex]

The limits of [latex]z[/latex] are [latex]{{r}^{2}}[/latex], hence

[latex]{\displaystyle\int^{{y} = {1}}_{{y} = {-1}}}\displaystyle\int^{{x} = {\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\displaystyle\int^{{z} = {\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \ {{x}{y}{z}} \ {{d}{z}} \ {{d}{x}} \ {{d}{y}} = {\displaystyle\int^{{\theta} = {{\pi}/{2}}}_{{\theta} = {-{\pi}/{2}}}} \ {\displaystyle\int^{{r} = {1}}_{{r} = {0}}} \ {\displaystyle\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \ {r}{({r}{\cos}{\theta})}{({r}{\sin}{\theta})}{z} \ {{d}{z}} \ {{d}{r}} \ {{d}{\theta}}[/latex].

The ranges of the variables are

[latex]\begin{aligned} {0} \ \ \ & {\leq} \ \ \ {y} \ {\leq} \ {3} \\ {0} \ \ \ & {\leq} \ \ \ {x} \ {\leq} \ {\sqrt{{9} - {{y}^{2}}}} \\ {\sqrt{{{x}^{2}} + {{y}^{2}}}} \ \ \ & {\leq} \ \ \ {z} \ {\leq} \ {\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}. \end{aligned}[/latex]

The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[/latex]-plane. Hence the range for [latex]{\theta}[/latex] is [latex]{0} \ {\leq} \ {\theta} \ {\leq} \ {\frac{\pi}{2}}[/latex].

The lower bound [latex]{z} = {\sqrt{{{x}^{2}} + {{y}^{2}}}}[/latex] is the upper half of a cone and the upper bound [latex]{z} = {\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}[/latex] is the upper half of a sphere. Therefore, we have [latex]{0} \ {\leq} \ {\rho} \ {\leq} \ {\sqrt{18}}[/latex], which is [latex]{0} \ {\leq} \ {\rho} \ {\leq} \ {3{\sqrt{2}}}[/latex].

For the ranges of [latex]{\varphi}[/latex], we need to find where the cone and the sphere intersect, so solve the equation

[latex]\begin{aligned} {{{r}^{2}} + {{z}^{2}}} \ \ \ & = \ \ \ {18} \\ {({\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \ \ \ & = \ \ \ {18} \\ {{z}^{2}} + {{z}^{2}} \ \ \ & = \ \ \ {18} \\ {{2}{{z}^{2}}} \ \ \ & = \ \ \ {18} \\ {{z}^{2}} \ \ \ & = \ \ \ {9} \\ {z} \ \ \ & = \ \ \ {3}. \end{aligned}[/latex]

This gives

[latex]\hspace{10cm}\begin{align} 3\sqrt2\cos\varphi&=3 \\ \cos\varphi&=\frac1{\sqrt2} \\ \varphi&=\frac{\pi}4. \end{align}[/latex]

Putting this together, we obtain

We calculate the volume of the ball in the first octant, where [latex]x\geq0[/latex], [latex]y\geq0[/latex], and [latex]z\geq0[/latex], using spherical coordinates, and then multiply the result by [latex]8[/latex] for symmetry. Since we consider the region [latex]D[/latex] as the first octant in the integral, the ranges of the variables are

[latex]\large{0\leq\varphi\leq\frac{\pi}2, \ 0\leq\rho\leq{r}, \ 0\leq\theta\leq\frac{\pi}2}[/latex]

Therefore,

[latex]\hspace{6cm}\begin{align} V&=\underset{D}{\displaystyle\iiint}dx \ dy \ dz = 8\displaystyle\int_{\theta=0}^{\theta=\pi/2}\displaystyle\int_{\rho=0}^{\rho=\pi}\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\ &=8\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}d\varphi\displaystyle\int_{\rho=0}^{\rho=\pi}\rho^2 \ d\rho\displaystyle\int_{\theta=0}^{\theta=\pi/2}\sin\theta \ d\theta \\ &=8\left(\frac{\pi}2\right)\left(\frac{r^3}3\right)(1) \\ &=\frac43\pi{r}^3 \end{align}[/latex]

This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[/latex] ft, the volume is [latex]\frac43\pi{r}^3\approx523,000\text{ ft.}^3[/latex].

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\geq0[/latex], [latex]y\geq0[/latex], and [latex]z\geq0[/latex] and then multiply the result by 8.

In this case the ranges of the variables are

[latex]\large{0\leq\varphi\leq\frac{\pi}2, \ 0\leq\rho\leq\frac{\pi}2, \ 0\leq\rho\leq1, \ 0\leq\theta\leq\frac{\pi}2}[/latex]

Also, we need to change the rectangular to spherical coordinates in this way:

[latex]\large{x=a\rho\cos\varphi\sin\theta, \ y=b\rho\sin\varphi\sin\theta,\text{ and }z=c\rho\cos\theta}.[/latex]

Then the volume of the ellipsoid becomes

[latex]\hspace{6cm}\begin{align}V&=\underset{D}{\displaystyle\iiint}dx \ dy \ dz \\&=8\displaystyle\int_{\theta=0}^{\theta=\pi/2}\displaystyle\int_{\rho=0}^{\rho=1}\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}abc\rho^2\sin\theta \ d\varphi \ d\rho \ d\theta \\&=8abc\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}d\varphi\displaystyle\int_{\rho=0}^{\rho=1}\rho^2 \ d\rho\displaystyle\int_{\theta=0}^{\theta=\pi/2}\sin\theta \ d\theta \\&=8abc\left(\frac{\pi}2\right)\left(\frac13\right)(1) \\&=\frac43\pi{abc}\end{align}[/latex]

This problem is directly related to the l’Hemisphèric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then subtract.

First we find the volume of the ellipsoid using [latex]a=75[/latex] ft, [latex]b=80[/latex] ft, and [latex]c=90[/latex] ft. The volume of the ellipsoid is:

[latex]\large{V_{\text{ellipsoid}}=\frac43\pi(75)(80)(90)\approx2,262,000\text{ ft.}^3}[/latex]

We found earlier, the volume of the sphere is:

[latex]\large{V_{\text{sphere}}\approx523,600\text{ ft.}^3.}[/latex]

Therefore, the volume of the space inside the ellipsoid [latex]\frac{x^2}{75^2}+\frac{y^2}{80^2}+\frac{z^2}{90^2}=1[/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[/latex] is approximately

[latex]\large{V_{\text{Hemisferic}}=V_{\text{ellipsoid}}-V_{\text{sphere}}=1,738,400\text{ ft.}^3.}[/latex]