• anonymous
Calculus question: surface area of a torus. it is generated by circle with radius r , off center. Suppose the circle is centered at (x - R)^2 + y^2 = r^2, and R > r. so the distance from (0,0) to the center of the circle is (R,0) which has radius r. Ok the integral i got is int 2pi y sqrt [ 1+ (dy/dx)^2] = int 2pi sqrt ( r^2 - ( x-R)^2) [ 1 + (-2(x-R)/(2sqrt( r^2 - ( x-R)^2) ^2 ]
Mathematics
• Stacey Warren - Expert brainly.com
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