Question

Use the result of Exercise 23 to set up an integral to find the area of the surface generated by rotating the curve \(y = \sqrt x ,0x4\) , about the line \(y = 4\). Then use a CAS to evaluate the integral.

Short answer

The result of wolfram alpha’s work will be 80.6095.

Step-by-step solution

Definition of work done

The area or region occupied by the object's surface is referred to as its surface area.

Finding the area of surface generated by rotating curve

In the last exercise, we discovered a formula that was:

\(S = \int_a^b 2 \pi [c - f(x)]\sqrt {1 + {{\left[ {{f^\prime }(x)} \right]}^2}} dx\)

In this case a=0,b=4,c=4 and f(x) = \(\sqrt 2 \)

\(\begin{aligned}S &= \int_a^b 2 \pi [c - f(x)]\sqrt {1 + {{\left[ {{f^\prime }(x)} \right]}^2}} dx\\S &= \int_0^4 2 \pi (4 - \sqrt x )\sqrt {1 + {{\left[ {\frac{{d(\sqrt x )}}{{dx}}} \right]}^2}} dx\\ &= \int_0^4 2 \pi (4 - \sqrt x )\sqrt {1 + {{\left[ {\frac{1}{{2\sqrt x }}} \right]}^2}} dx\\ &= \int_0^4 2 \pi (4 - \sqrt x )\sqrt {1 + \frac{1}{{4x}}} dx\end{aligned}\)

To solve this on wolfram alpha (which can be used as CAS), consider the following text: integrate \(2\backslash pi(4 - \sqrt x )\sqrt {1 + \frac{1}{{4x}}} \) on [0, 4]

The result of wolfram alpha’s work will be 80.6095.