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Polar coordinates are particularly useful for problems involving symmetry around a point, often the origin of our coordinate system.
Instead of using the usual x-y coordinate grid, polar coordinates describe a point in space using a distance from the origin, r, and an angle from a reference direction, \( \theta \). This makes it easier to represent circles and radial boundaries.
In this exercise, the surface equations are given in terms of r and \( \theta \), making polar coordinates a natural choice for simplifying the problem.

• Polar coordinates express x and y in terms of r and \( \theta \):
• \( x = r \cos \theta \)
• \( y = r \sin \theta \)

Understanding these transformations helps in visualizing and setting boundaries for solids in three-dimensional space, like the task at hand.

Triple integration is the method of integrating a function of three variables over a three-dimensional region.
It is an extension of double integration, which is used for area, to compute volume by adding a third variable for height. In this exercise, however, we determine the volume in the first octant using a double integral because the function is already simplified in polar coordinates.The integral is split into inner and outer integrals, with the inner integral often involving the radius r, and the outer concerning the angle \( \theta \). For this problem:

• The inner integral computes the volume over varying \( r \): \( \int_0^{3 \sin \theta} r^2 \sin \theta \, dr \)
• The outer integral sums these slices over all \( \theta \): \( \int_0^{\pi/2} 9 \sin^4 \theta \, d\theta \)

This method simplifies calculating volumes by breaking the problem into manageable parts. The calculations involve substituting expressions for z, in terms of polar coordinates, into the integral to properly define the region of interest.

The first octant refers to the region in 3D space where all three coordinates, x, y, and z, are non-negative.
This confines the solid to one eighth of the entire space around the origin, where typically, geometrical figures are more easily analyzed.
For this problem, only points where \( r \geq 0 \), \( \theta \geq 0 \) and \( z \geq 0 \) are considered, simplifying the bounds of integration.In the first octant, \( \theta \) ranges from 0 to \( \frac{\pi}{2} \), which corresponds to one quarter of a full circle, effectively limiting the volume to the positive x and y coordinates.

• Bound by \( x = 0 \) and \( y = 0 \)
• Generally, also by \( z = 0 \) at the bottom and some surface on top

This spatial restriction helps to pinpoint exact volumes with fewer calculations, as the symmetry and positive axes limit possibilities.