Question

A pharmaceutical company is designing a new drug whose shape in tablet form is obtained by rotating the graph of ${\left.f\left(x\right)=2\left(4\right)-x^2\right)}^{1/4}$on the interval $[-6,6]$around the x-axis, where units are measured in millimeters.

As we will see in Section 6.1, the volume of such a solid is given by $\pi {\int }_a^b\left(f\right(x){)}^{2}dx$, where a and b are the places where the graph of $f\left(x\right)$meets the x-axis.

(a) Write down a specific definite integral that represents the amount of material, in cubic millimeters, required to make the tablet.

(b) Use trigonometric substitution to solve the definite integral and determine the amount of material needed.

Short answer

The answer is,

Part (a) The integral is $4\pi {\int }_{-\sqrt{42}}^{\sqrt{42}}\sqrt{42-x^2}dx$

Part (b) The value is$829{\mathrm{mm}}^3$

Step-by-step solution

Part (a) Step 1. Given Information.

The given function is$f\left(x\right)=2{\left(42-x^2\right)}^{1/4}\text{on the interval}[-6,6]$.

Part (a) Step 2. Explanation.

We know,

Volume is given by $\pi {\int }_a^b\left(f\right(x){)}^{2}dx$

$$\begin{gathered} =\pi {\int }_{-6}^6{\left(2{\left(42-x^2\right)}^{\frac{1}{4}}\right)}^{2}dx \\ =4\pi {\int }_{-\sqrt{42}}^{\sqrt{42}}\sqrt{42-x^2}dx \end{gathered}$$

Part (b) Step 1. Explanation.

On using trigonometric substitution,

$$\begin{gathered} 4\pi {\int }_{-\sqrt{42}}^{\sqrt{42}}\sqrt{42-x^2}dx=4\pi {\int }_{-\sqrt{42}}^{\sqrt{42}}\sqrt{42-(\sqrt{42}\sin u{)}^2}du \\ =829 \end{gathered}$$