Question

Evaluate (x2 + y2) dv e , where e lies between the spheres x2 + y2 + z2 = 9 and x2 + y2 + z2 = 16.

Answer 1

Convert to spherical coordinates, using

$x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi$
$y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi$
$z(\rho,\theta,\varphi)=\rho\cos\varphi$

The volume element is

$\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

In spherical coordinates, the two given spheres are obtained by setting $\rho=3$ and $\rho=4$. So the integral evaluates to

$\displaystyle\iiint_{\mathcal E}(x^2+y^2)\,\mathrm dV=\int_{\varphi=0}^{\varphi=\pi}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=3}^{\rho=4}\rho^2\sin^2\varphi(\rho^2\sin\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$
$=\dfrac{6248\pi}{15}$

Answer 2

The volume between the spheres is given by the triple integration of (x² + y²) is 6248π/15.

What is an area bounded by the curve?

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The spheres x² + y² + z² = 9 and x² + y² + z² = 16.

Convet into shperical coordinate

$x(\rho , \theta , \phi) = \rho \cos \theta \sin \phi\\\\y \rho , \theta , \phi) = \rho \sin \theta \cos \phi \\\\z(\rho , \theta , \phi) = \rho \cos \phi$

Then the volume of the element will be

$\rm dV = \rho ^2 \sin \phi \ d \rho \ \d\theta \ d\phi$

Then we have

$\rightarrow \int \int \int_{\epsilon} (x^2 + y^2 ) dV \\\\\\\rightarrow \int _{\phi = 0}^{\phi = \pi} \int _{\theta = 0}^{\theta = 2\pi} \int _{\rho = 3}^{\rho = 4} \rho ^2 \sin^2 \phi (\rho ^2 \sin \phi) d\rho \ d\theta \ d\phi\\\\\\\rightarrow \dfrac{6248 \pi}{15}$

More about the area bounded by the curve link is given below.

https://brainly.com/question/24563834