A. Let B be the region between the upper concentric...

a. Let $B$ be the region between the upper concentric hemispheres of radii $a$ and $b$ centered at the origin and situated in the first octant, where $0 < a < b$.spherical coordinates $\quad(\rho, \theta, \varphi) \qquad$ is $F(x, y, z)=f(\rho) \cos \varphi . \quad$ Show that $\quad$ if $$ g(a)=g(b)=0 \text { and } \int_{a}^{b} h(\rho) d \rho=0, $$ $$ \iiint_{B} F(x, y, z) d V=\frac{\pi^{2}}{4}[a h(a)-b h(b)] $$ where $g$ is an antiderivative of $f$ and $h$ is an antiderivative of $g .$ b. Use the previous result to show that $$ \iiint_{B} \frac{z \cos \sqrt{x^{2}+y^{2}+z^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}} d V=\frac{3 \pi^{2}}{2}, \text { where } B \text { is } $$ the region between the upper concentric hemispheres of radii $\pi$ and 2$\pi$ centered at the origin and situated in the first octant.

a. Let $B$ be the region between the upper concentric hemispheres of radii $a$ and $b$ centered at the origin and situated in the first octant, where $0 < a < b$.spherical coordinates $\quad(\rho, \theta, \varphi) \qquad$ is $F(x, y, z)=f(\rho) \cos \varphi . \quad$ Show that $\quad$ if
$$
g(a)=g(b)=0 \text { and } \int_{a}^{b} h(\rho) d \rho=0,
$$
$$
\iiint_{B} F(x, y, z) d V=\frac{\pi^{2}}{4}[a h(a)-b h(b)]
$$
where $g$ is an antiderivative of $f$ and $h$ is an antiderivative of $g .$
b. Use the previous result to show that
$$
\iiint_{B} \frac{z \cos \sqrt{x^{2}+y^{2}+z^{2}}}{\sqrt{x^{2}+y^{2}+z^{2}}} d V=\frac{3 \pi^{2}}{2}, \text { where } B \text { is }
$$
the region between the upper concentric hemispheres of radii $\pi$ and 2$\pi$ centered at the origin and situated in the first octant.

Key Concepts

Spherical Coordinates

Spherical coordinates provide an alternative way to represent points in three-dimensional space using a radial distance and two angles. They are particularly useful when the region of integration or the function being integrated has spherical symmetry, as they can simplify the bounds and the expression of the integrand.

Jacobian Determinant in Spherical Coordinates

When converting an integral from Cartesian to spherical coordinates, the volume element changes by the Jacobian determinant. In spherical coordinates, the differential volume element is ?² sin ? d? d? d?, which accounts for the stretching and compression of space in the new coordinate system.

Triple Integration

Triple integration is the process of integrating a function over a three-dimensional region. It involves integrating sequentially with respect to each variable, and in cases where symmetry is present, an appropriate coordinate system (like spherical coordinates) is often used to simplify the integration process.

Region of Integration

The region of integration is the specific portion of space over which the integral is evaluated. In problems involving spherical coordinates, regions defined between spheres or hemispheres and restricted to certain octants can be expressed with simple bounds in ?, ?, and ?, making use of the geometric features of the region.

Antiderivatives and Their Role in Integration

Antiderivatives are functions whose derivatives yield the original function, and they are crucial in evaluating definite integrals via the Fundamental Theorem of Calculus. In the context of the given problem, specifying antiderivative boundary conditions and using the fact that the integral of a derivative over an interval produces a difference in antiderivative values is central to obtaining the final result.

Transcript

00:01 In this question, we are given an integral and we want to prove a certain result.

00:06 So let's first work on the integral.

00:09 We know that the limits of b are hemispheres with radii a and b, and we also know that it is situated in the first octon, and so that means that the limits of theta and phi would be 0 to pi divided by 2 because we only want the first quarter of each plane and the limits of row would be from a to b.

00:35 Next, we can separate this complicated integral into three separate parts because none of the limits depend on each other as in they don't depend on other variables.

00:49 So we can work on the integrals for with respect to theta and phi.

00:55 We can use trigonometric identities to work on cosine 5 sign 5 and we would be able to after some simplification we would be able to get the results after integrating with respect to phi and theta and we are now left with a pretty complicated we are now left with the integral with respect to roe so back to the question we know that when we integrate f, we get g and when we integrate function g, we get function h.

01:35 So using integration by parts, i can rewrite the last integral with respect to row.

01:44 As such, we can eliminate the first term.

01:49 So, row squared times g, row.

01:53 We can eliminate that because in the question it says that g of a is equal to g of b which is equal to zero so that term would become zero now we have an integral with now we have an integral with respect to role and the function we want to integrate is g again we can apply integration by parts and express this integral as such so we can simplify it again because when we integrate this we have 2b h of b minus 2a h of a so we can so we can replace this result this the result of this integral to back to our question so we have half from the integral with respect to phi pi divided by 2 from the integral with respect to theta and then we can substitute the in the result of the integral we just found back into what we have so far and that gives us the result that we want to prove in the question next we want to use a previous result to show something so let us first look at the limits of b so again b is just a as in the previous case, the limits of b are hemispheres in the first octon, and we have the values of a and b now.

03:35 The value of a would be pi, and the value of b would be 2 pi.

03:41 And we can rewrite our function, the function that we want to integrate with this, which is given in the question...

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