Evaluate the integral below, where e lies between the spheres x2 + y2 + z2 = 9 and x2 + y2 + z2 = 25 in the first octant.

The student's question involves integrating a function in a region bounded by two spheres in the first octant, implying the use of spherical coordinates and integration over a sphere with a constant radius.

The question pertains to evaluating an integral within the region bounded by two spheres in the first octant. When dealing with spheres and integrals, the use of spherical coordinates is often beneficial. The question suggests using spheres with a constant radius and spherical coordinates (r, θ, φ), where a typical point in space is represented as (r sin(θ) cos(φ), r sin(θ) sin(φ), r cos(θ)). To integrate over the sphere, we consider the bounds given by the radii of the inner and outer spheres, (r = 3 and r = 5, respectively, since the square roots of 9 and 25 are 3 and 5), and the fact that it is within the first octant which further restricts the limits of θ and φ. The rest of the provided excerpts seem to be unrelated specifically to this problem but are examples of standard integrals and applications of integration in physics and potential theory.

The final answer after evaluating the integral is: [tex]\[\frac{49\pi}{3}\][/tex]. This is the value of the integral over the region between the spheres [tex]\( x^2 + y^2 + z^2 = 9 \) and \( x^2 + y^2 + z^2 = 25 \)[/tex] in the first octant.

To evaluate the given integral over the region between the spheres [tex]\( x^2 + y^2 + z^2 = 9 \)[/tex]and [tex]\( x^2 + y^2 + z^2 = 25 \)[/tex]  in the first octant, we can use spherical coordinates. In spherical coordinates, the volume element is given by [tex]\( r^2 \sin(\phi) \, dr \, d\theta \, d\phi \),[/tex] where r is the radial distance, [tex]\( \theta \)[/tex] is the azimuthal angle, and [tex]\( \phi \)[/tex] is the polar angle.

The limits for the integral are as follows:

[tex]- \( 3 \leq r \leq 5 \) (limits of the radii for the spheres)\\- \( 0 \leq \theta \leq \frac{\pi}{2} \) (first octant)\\- \( 0 \leq \phi \leq \frac{\pi}{2} \) (first octant)[/tex]

The integral to evaluate is not specified, so let's assume it's a simple function like \( f(x, y, z) = 1 \) for the sake of demonstration. The integral would then be:

[tex]\[\iiint_E 1 \, dV = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{3}^{5} r^2 \sin(\phi) \, dr \, d\theta \, d\phi\][/tex]

Now, let's evaluate this integral step by step:

[tex]\[\int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \int_{3}^{5} r^2 \sin(\phi) \, dr \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left[ \frac{1}{3} r^3 \sin(\phi) \right]_{3}^{5} \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \left( \frac{125}{3} - \frac{27}{3} \right) \sin(\phi) \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \int_{0}^{\frac{\pi}{2}} \frac{98}{3} \sin(\phi) \, d\theta \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \left[ \frac{98}{3} \theta \right]_{0}^{\frac{\pi}{2}} \, d\phi\][/tex]

[tex]\[= \int_{0}^{\frac{\pi}{2}} \frac{98}{3} \cdot \frac{\pi}{2} \, d\phi\][/tex]

[tex]\[= \frac{98\pi}{6}\][/tex]

[tex]\[= \frac{49\pi}{3}\][/tex]

So, the value of the integral over the specified region is[tex]\( \frac{49\pi}{3} \).[/tex]

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