Question

Choose the best coordinate system and find the volume of the following solid regions. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The region inside the sphere \(\rho=1\) and below the cone \(\varphi=\pi / 4\), for \(z \geq 0\)

Short answer

Question: Find the volume of a solid region inside a sphere with ρ = 1 and below a cone with φ = π/4, for z≥0. Use spherical coordinates to compute the volume. Answer: The volume of the given solid region is V = (2π/3)(1 - √2/2).

Step-by-step solution

Determine the limits of integration

The region is inside the sphere with \(\rho=1\). Therefore, the limits for \(\rho\) are: $$ 0 \leq \rho \leq 1 $$ The cone is defined by \(\varphi=\pi/4\). Since we're considering \(z\geq 0\), the limits for \(\varphi\) are: $$ 0 \leq \varphi \leq \frac{\pi}{4} $$ Finally, the variable \(\theta\) covers the entire circle, so its limits are: $$ 0 \leq \theta \leq 2\pi $$

Write out the triple integral

Now that we have determined the limits of integration, we can write out the triple integral representing the volume of the solid region: $$ V = \int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}} \rho^2 \sin{\varphi}\,d\rho\,d\theta\,d\varphi $$

Integrate with respect to \(\rho\)

First, integrate with respect to \(\rho\): $$ V = \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}} \frac{1}{3} \rho^3 \sin{\varphi} \Big|_{0}^{1}\,d\theta\,d\varphi = \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}} \frac{1}{3} \sin{\varphi} \,d\theta\,d\varphi $$

Integrate with respect to \(\theta\)

Next, integrate with respect to \(\theta\): $$ V = \int_{0}^{\frac{\pi}{4}} \frac{1}{3} \sin{\varphi} (\theta \Big|_{0}^{2\pi})\,d\varphi = \int_{0}^{\frac{\pi}{4}} \frac{2\pi}{3} \sin{\varphi} \,d\varphi $$

Integrate with respect to \(\varphi\)

Finally, integrate with respect to \(\varphi\): $$ V = \frac{2\pi}{3} (-\cos{\varphi} \Big|_{0}^{\frac{\pi}{4}}) = \frac{2\pi}{3}(1 - \frac{\sqrt{2}}{2}) $$ Hence, the volume of the given solid region is: $$ V = \frac{2\pi}{3}(1 - \frac{\sqrt{2}}{2}) $$

Key concepts

Triple Integrals

Triple integrals are a fundamental tool in solid geometry. They allow you to find the volume of a region in three-dimensional space. A triple integral involves three iterations of integration, one for each of the three dimensions. In the context of cylindrical or spherical coordinates, each integral corresponds to one of the coordinate axes. The order of integration can often be chosen freely, which can simplify the calculation.
For example, in this exercise, we're dealing with a solid inside a sphere and below a cone. Establishing the limits of integration for each variable is crucial:

• First, we integrate with respect to the radial coordinate, \(\rho\), because it varies from the center to the boundary of the sphere.
• Second, we handle the azimuthal angle, \(\theta\), which spans the complete circle, simplifying our calculation.
• Finally, for the vertical angle, \(\varphi\), we integrate from the top of the cone to the base of the hemisphere.

Each integration step refines our calculation of the region's volume. Getting these limits right is key to ensuring our use of the triple integral is correct.

Polar Coordinates

Polar coordinates are useful when dealing with circular and spherical shapes because they simplify the equations representing curves or surfaces. Unlike Cartesian coordinates, which define a point by x, y, and z distances from the origin, polar coordinates use angles and radii.
In this exercise, we use spherical coordinates, an extension of polar coordinates, because of the symmetry in the problem. Spherical coordinates consist of three parameters:

• The radial distance, \(\rho\), from the origin to the point.
• The polar angle, \(\varphi\), measured from the positive z-axis.
• The azimuthal angle, \(\theta\), within the x-y plane from the positive x-axis.

These coordinates are perfect for a sphere and a cone since they match the natural shape of our region.
Using spherical coordinates, the volume integrals become much easier to handle, paving the way for more straightforward calculations.

Volume of Solids

The volume of solids is often calculated using triple integrals, especially when the regions are bounded by complex surfaces like spheres and cones. Understanding how to set appropriate limits for these integrals in spherical coordinates can simplify finding the volume.
In this context, solving for volume involves:

• Setting up the triple integral within the context of the solid's boundaries, using the simplest coordinate representation.
• Evaluating these integrals by integrating with respect to each variable sequentially.
• Taking advantage of symmetry when possible, as symmetrical regions allow for simpler integration.

For instance, the exercise provided us with a solid bound by a sphere and a cone. A successful breakdown involves first defining the exact limits for \(\rho\), \(\theta\), and \(\varphi\). By understanding the geometry, you translate these natural limits into an integrable form.
Finally, carrying through with the integration yields the solid's volume, allowing us to quantify the three-dimensional space it occupies.