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In spherical coordinates, we set

[tex]x=\rho\cos\theta\sin\varphi[/tex]

[tex]y=\rho\sin\theta\sin\varphi[/tex]

[tex]z=\rho\cos\varphi[/tex]

so that the volume element under this transformation becomes

[tex]\mathrm dV=\mathrm dx\,\mathrm dy\,\mathrm dz=|\det\mathbf J|\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]

The region [tex]E[/tex] is given by the set

[tex]\left\{(\rho,\theta,\varphi)\mid0\le\rho\le1,0\le\theta\le2\pi,0\le\varphi\le\dfrac\pi3\right\}[/tex]

so that the integral is

[tex]\displaystyle\iiint_Ey^2z^2\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi/3}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^6\sin^2\theta\sin^3\varphi\cos^2\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi[/tex]

[tex]\displaystyle=\left(\int_0^{\pi/3}\sin^3\varphi\cos^2\varphi\,\mathrm d\varphi\right)\left(\int_0^{2\pi}\sin^2\theta\,\mathrm d\theta\right)\left(\int_0^1\rho^6\,\mathrm d\rho\right)[/tex]

[tex]=\dfrac{47}{480}\cdot\pi\cdot\dfrac17=\dfrac{47\pi}{3360}[/tex]

ajeigbeibraheem

The volume of the solid using the triple integral [tex]\mathbf{\iiint_E y^2z^2 dV \ \ is \ \ = \dfrac{47 \pi}{3360}}[/tex]

Given that;

  • The solid E lies above the cone [tex]\mathbf{\pi = \dfrac{\pi}{3}}[/tex] as well as;
  • Below the sphere [tex]\mathbf{\rho = 1}[/tex]  

Then the spherical coordinates can be expressed as:

  • [tex]\mathbf{\rho^2 =x^2 +y^2 +z ^2}[/tex]

where;

  • [tex]\mathbf{x = \rho \ sin \phi \ cos \theta}[/tex]
  • [tex]\mathbf{y= \rho \ sin \phi \ sin \theta}[/tex]
  • [tex]\mathbf{z = \rho\ cos \theta}[/tex]  
  • [tex]\mathbf{dxdydz = \rho^2 \ sin \phi \ d \rho\ d \phi \ d \theta}[/tex]

Now, the expression for the solid E in the spherical coordinates can be computed as:

[tex]\mathbf{E = \Big \{ ( \rho , \theta, \phi )\Big| 0 \leq \rho \leq 1, 0 \leq \phi \leq \dfrac{\pi}{3}, 0 \leq \theta \leq 2 \pi \Big \} }[/tex]

and the volume of the solid using the triple integral is calculated as:

[tex]\mathbf{\iiint_E y^2z^2 dV = \iiint _E \ y^2 z^2 \ dx dy dz }[/tex]

[tex]\mathbf{\implies \iiint _E \ y^2 z^2 \ dx dy dz = \int ^{2 \pi}_{0} \int ^{\dfrac{\pi}{3}}_{0} \int ^1_0 \ (\rho sin \phi sin \theta )^2 ( \rho cos \phi )^2 \rho^2 sin \phi d \rhod \phi d \theta }[/tex]

[tex]\mathbf{\implies \int ^{2 \pi}_{0} \int ^{\dfrac{\pi}{3}}_{0} \ sin^3 \phi cos^2 \phi sin^2 \theta \ \int^1_0 \ \rho^6 d \rho d \phi d \theta }[/tex]

[tex]\mathbf{\implies \int ^{2 \pi}_{0} \int ^{\dfrac{\pi}{3}}_{0} \ sin^3 \phi cos^2 \phi sin^2 \theta \Big [ \dfrac{\rho^7}{7} \Big]^1_0 \ d \phi d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta \int ^{\dfrac{\pi}{3}}_{0} \ sin^3 \phi cos^2 \phi \ d \phi d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta \int ^{\dfrac{\pi}{3}}_{0} \ sin \phi( 1- cos^2 \phi)cos^2 \phi \ d \phi d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta\Bigg [\dfrac{cos^5 \phi}{5}- \dfrac{cos ^3 \phi}{3} \Bigg ] ^{\dfrac{\pi}{3}}_{0} d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta\Bigg [\dfrac{cos^5 \dfrac{\pi}{3}}{5}- \dfrac{cos ^3 \dfrac{\pi}{3}}{3}- \dfrac{cos^5 0}{5}+ \dfrac{cos^3 0}{3} \Bigg ] d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta\Bigg [\dfrac{2}{15}- \dfrac{17}{480} \Bigg ] d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta\Bigg [\dfrac{(480\times 2) -(15\times 17)}{15\times 480}\Bigg ] d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{1}{7} \int ^{2 \pi}_{0} sin ^2 \theta\Bigg [\dfrac{705}{15\times 480}\Bigg ] d \theta }[/tex]

[tex]\mathbf{\implies \dfrac{47}{6720} \int ^{2 \pi}_{0} 2sin ^2 d \theta}[/tex]

[tex]\mathbf{\implies \dfrac{47}{6720} \int ^{2 \pi}_{0} (1-cos 2\theta) \ d \theta}[/tex]

[tex]\mathbf{\implies \dfrac{47}{6720} \Bigg [\theta - \dfrac{sin 2 \theta }{2}\Bigg] ^{2 \pi}_{0}}[/tex]

[tex]\mathbf{\implies \dfrac{47}{6720} \Bigg [2 \pi\Bigg] }[/tex]

[tex]\mathbf{\iiint_E y^2z^2 dV = \dfrac{47 \times 2 \pi}{6720}} }[/tex]

[tex]\mathbf{\iiint_E y^2z^2 dV = \dfrac{47 \pi}{3360} }[/tex]

Learn more about triple integral here:

https://brainly.com/question/2289273?referrer=searchResults

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