Calculus II - Surface Area (Assignment Problems)

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Section 8.2 : Surface Area

Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating y=7x+2 , −5≤y≤0 about the x-axis using,
1. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$
2. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$
Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating y=1+2x5 , 0≤x≤1 about the x-axis using,
1. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$
2. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$
Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating x=e2y , −1≤y≤0 about the y-axis using,
1. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx$
2. $\displaystyle ds = \sqrt {1 + {{\left[ {\frac{{dx}}{{dy}}} \right]}^2}} \,dy$
Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating y=cos(12x) , 0≤x≤π about
1. the $x$ -axis
2. the $y$ -axis
Set up, but do not evaluate, an integral for the surface area of the object obtained by rotating x=√3+7y , 0≤y≤1 about
1. the $x$ -axis
2. the $y$ -axis
Find the surface area of the object obtained by rotating $\displaystyle y = \frac{1}{4}\sqrt {6x + 2}$ , $\displaystyle \frac{{\sqrt 2 }}{2} \le y \le \frac{{\sqrt 5 }}{2}$ about the $x$ -axis.
Find the surface area of the object obtained by rotating $y = 4 - x$ , $1 \le x \le 6$ about the $y$ -axis.
Find the surface area of the object obtained by rotating $x = 2y + 5$ , $- 1 \le x \le 2$ about the $y$ -axis.
Find the surface area of the object obtained by rotating $x = 1 - {y^2}$ , $0 \le y \le 3$ about the $x$ -axis.
Find the surface area of the object obtained by rotating $x = {{\bf{e}}^{2y}}$ , $- 1 \le y \le 0$ about the $y$ -axis.
Find for the surface area of the object obtained by rotating $y = \cos \left( {\displaystyle \frac{1}{2}x} \right)$ , $0 \le x \le \pi$ about the $x$ -axis.