Respuesta :
The volume is given by the integral,
[tex]\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]
That is, [tex]\rho[/tex] ranges from the origin to the sphere of radius 4. The range for [tex]\varphi[/tex] starts at the intersection of the cone [tex]z=x^2+y^2[/tex] with the sphere [tex]x^2+y^2+z^2=16[/tex], which gives
[tex]z+z^2=16\implies z^2+z-16=0\implies z=\dfrac{\sqrt{65}-1}2[/tex]
and
[tex]z=4\cos\varphi\implies\varphi=\cos^{-1}\left(\dfrac{\sqrt{65}-1}8\right)[/tex]
and extends to the x-y plane where [tex]\varphi=\frac\pi2[/tex]. The range for [tex]\theta[/tex] is self-evident.
The volume is then
[tex]V=\displaystyle\int_0^{2\pi}\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\int_0^4\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta[/tex]
[tex]V=\displaystyle\left(\int_0^{2\pi}\mathrm d\theta\right)\left(\int_0^{\cos^{-1}((\sqrt{65}-1)/8)}\sin\varphi\,\mathrm d\varphi\right)\left(\int_0^4\rho^2\,\mathrm d\rho\right)[/tex]
[tex]V=2\pi\left(\dfrac{\sqrt{65}-1}8\right)\left(\dfrac{64}3\right)=\boxed{\dfrac{16\pi(9-\sqrt{65})}3}[/tex]
A sphere is a three-dimensional object with a round form. The volume of the sphere is [16π(9-√65)]/3 unit³.
What is a sphere?
A sphere is a three-dimensional object with a round form. A sphere, unlike other three-dimensional shapes, has no vertices or edges. Its centre is equidistant from all places on its surface. In other words, the distance between the sphere's centre and any point on its surface is the same.
We know that the volume of the given sphere can be given by the integral,
[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta[/tex]
where ρ ranges from the origin of the plot to the sphere of radius 4 while the range of φ starts at the intersection of the cone z=x²+y² with the sphere x²+y²+z²=16.
Now, the value of z and φ can be written as,
[tex]x^2+y^2+z^2 = 16\\\\(x^2+y^2)+z^2 = 16\\\\z+z^2 = 16\\\\z^2+z-16=0 \implies z=\dfrac{\sqrt{65}-1}{2}[/tex]
And
[tex]z =4\ cos\ \varphi \implies \varphi =cos^{-1}(\dfrac{\sqrt{65}-1}{8})[/tex]
Further, the volume of the sphere can be written as,
[tex]{\rm Volume} = \int^{2\pi}_0\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} \int_0^4\rho^2sin\varphi\ d\rho\ d\varphi\ d\theta\\\\\\{\rm Volume} = (\int^{2\pi}_0\ d\theta)(\int^{cos^{-1}(\frac{\sqrt{65}-1}{8})} sin\varphi\ d\varphi)(\int_0^4\rho^2 d\rho)\\\\\\V = 2\pi(\dfrac{\sqrt{65}-1}{8})(\dfrac{64}{3}) = \dfrac{16\pi(9-\sqrt{65})}{3}[/tex]
Hence, the volume of the sphere is [tex]\dfrac{16\pi(9-\sqrt{65})}{3}[/tex].
Learn more about Sphere:
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