Paul's Online Notes
Home / Calculus II / Applications of Integrals / Surface Area
Show All Notes Hide All Notes

Section 8.2 : 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