Question

Consistency of Volume Definitions The volume formulas in calculus are consistent with the standard formulas from geometry in the sense that they agree on objects to which both apply. (a) As a case in point, show that if you revolve the region enclosed by the semicircle \(y=\sqrt{a^{2}-x^{2}}\) and the \(x\) -axis about the \(x\) -axis to generate a solid sphere, the calculus formula for volume at the beginning of the section will give \((4 / 3) \pi a^{3}\) for the volume just as it should. (b) Use calculus to find the volume of a right circular cone of height \(h\) and base radius \(r .\)

Short answer

The results are consistent with the geometric formulas for the volumes of a sphere and a cone, proving that the calculus formulation for volumes (\(4/3 \pi a^{3}\) and \(1/3 \pi r^{2}h\), respectively) aligns with the geometric understanding of these figures.

Step-by-step solution

Use the Disk Method to Compute the Volume of a Solid Sphere

The Disk Method involves integrating the area of a circular slice, or disk, along the axis of revolution (in this case, the x-axis). The equation for the semicircle revolved around the x-axis is \(y=\sqrt{a^{2}-x^{2}}\), and the area of each disc at position x with thickness dx is \(A(x) = \pi (\sqrt{a^{2}-x^{2}})^{2} dx = \pi (a^{2}-x^{2}) dx\). Integrating this area from -a to a will give us the volume of the sphere: \(V = \int_{-a}^{a} A(x) dx = \pi \int_{-a}^{a} (a^{2}-x^{2}) dx\).

Solve the Integral for the Sphere

The specific integrand for the volume of the sphere can be solved using basic integral calculus rules. The resulting integral equals \(\pi [a^{2}x - (1/3)x^{3} ]_{-a}^{a} = (4/3) \pi a^{3}\). The volume of the sphere computed using calculus thus agrees with the geometric formula.

Use the Disk Method to Compute the Volume of a Cone

For a right circular cone with height h and base radius r, consider it as a solid of revolution of the line \(y = rh/x\) around the x-axis. The area of the disc at position x with thickness dx here is \(A(x) = \pi (rh/x)^{2} dx = \pi r^{2}h^{2}/x^{2} dx\). We integrate this area from h to 0 to get the volume: \(V=\int_{h}^{0} A(x)dx = \pi \int_{h}^{0} r^{2}h^{2}/x^{2} dx\).

Solve the Integral for the Cone

Again, the specific integral for the cone volume can be solved using basic integral calculus rules, to get: \(V = \pi h^{2} r^{2} [x^{-1}]_{h}^{0} = 1/3 \pi r^{2} h\). Calculating the volume with the Disk Method validates the well-known formula for the volume of a cone.