Provide an Appropriate Response $r = f ( \theta )$ From $\theta = \alpha$

Provide an appropriate response.
-The area of the surface formed by revolving the graph of $r = f ( \theta )$ from $\theta = \alpha$ to $\theta = \beta$ about the line $\theta = 0$ (the polar axis) is $\mathrm { SA } = \int _ { \alpha } ^ { \beta } 2 \pi \mathrm { r } \sin \theta \sqrt { \mathrm { r } ^ { 2 } + \left( \frac { \mathrm { dr } } { \mathrm { d } \theta } \right) ^ { 2 } } \mathrm {~d} \theta$ . The area of the surface formed by revolving the graph of $\mathrm { r } = \mathrm { f } ( \theta )$ from $\theta = \alpha$ to $\theta = \beta$ about the line $\theta = \frac { \pi } { 2 }$ is $\mathrm { SA } = \int _ { \alpha } ^ { \beta } 2 \pi \mathrm { r } \cos \theta \sqrt { \mathrm { r } ^ { 2 } + \left( \frac { \mathrm { dr } } { \mathrm { d } \theta } \right) ^ { 2 } } \mathrm {~d} \theta$ . Use the idea underlying these two integral formulas to find the surface area of the torus formed by revolving the circle $r =$ a about the line $r = b \sec \theta$ where $0 < a < b$ .