• anonymous
Surface area of a torus which has radius r which is generated by the circle centered at (R,0) with radius r. And assume R>r. So the equation of the circle is (x-R)^2 +y^2 = r^2. FOr some reason my integral does not work. i have integral 2pi y dS, where y = sqrt [r^2 - (x-R)^2], and ds = sqrt ( 1 + dy/dx ^2). dy/dx = -2(x-R)/ (2sqrt(r^2 - (x-R)^2)). anyways, integration limits , the lower limit is R-r, and upper R+r
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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