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May 6, 2021

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Section 2-2 : Surface Area

1. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$x = \sqrt {y + 5}$$ , $$\sqrt 5 \le x \le 3$$ about the $$y$$-axis using,
1. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$$
2. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$$
Solution
2. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$y = \sin \left( {2x} \right)$$ , $$\displaystyle 0 \le x \le \frac{\pi }{8}$$ about the $$x$$-axis using,
1. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$$
2. $$\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$$
Solution
3. Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating $$y = {x^3} + 4$$ , $$1 \le x \le 5$$ about the given axis. You can use either ds.
1. the $$x$$-axis
2. the $$y$$-axis
Solution
4. Find the surface area of the object obtained by rotating $$y = 4 + 3{x^2}$$ , $$1 \le x \le 2$$ about the $$y$$-axis. Solution
5. Find the surface area of the object obtained by rotating $$y = \sin \left( {2x} \right)$$ , $$\displaystyle 0 \le x \le \frac{\pi }{8}$$ about the $$x$$-axis. Solution