Question

The surfaces \(\rho = 1 + \frac{1}{5}sin m\theta sin n\theta \) have been used as models for tumors. The "bumpy sphere" with \(m = 6\)and \(n = 5\) is shown. Use a computer algebra system to find the volume it encloses.

Short answer

The volume of the bumpy sphere is \({\rm{ Volume }} = \frac{{136\pi }}{{99}}\)

Step-by-step solution

Define the Bumpy sphere.

In applied mathematics, the "bumpy sphere" with an equation in spherical coordinates, where and are positive values and and are positive integers, can be used to describe tumors progression. Demonstrate that the "bumpy sphere" is contained within an equation sphere.

Calculate the values of and at the point where the two surfaces meet.

Find the volume of bumpy sphere

Equation for the bumpy sphere is \(\rho = 1 + \frac{1}{5}\sin 6\theta \cdot \sin 5\phi \)

Define the region inside the bumpy sphere as

\begin{align} & E=\left\{ (\rho ,\theta ,\phi )\mid 0\le \rho \le 1+\frac{1}{5}\sin 6\theta \cdot \sin 5\phi ,\ \ \ 0\le \theta \le 2\pi ,\ \ \ 0\le \phi \le \pi \right\} \\ & \text{Volume }=\iiint_{E}{d}V =\int_{0}^{\pi }{\int_{0}^{2\pi }{\int_{0}^{1+\frac{1}{5}\sin 6\theta \cdot \sin 5\phi }{{{\rho }^{2}}}}}\sin \phi d\rho d\theta d\phi \\ &=\frac{136\pi }{99} \end{align}

Thus, the volume of the bumpy sphere is \({\rm{ Volume }} = \frac{{136\pi }}{{99}}\).