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Find out the surface area of the solid acquired by rotating y = √ (9-x^{2}), - 2 < x < 2 about the x-axis.
Solution
The formula that we'll be using here is,
S = ∫ 2Πyds
As we are rotating about the x-axis and we will make use of the first ds in this case since our function is in the correct form for this reason ds and we won't gain anything by solving it for x. Let us first get the derivative and the root taken care of.
Dy/dx = ½ (9-x^{2})^{- ½} (-2x)
= - x / (9-x^{2})^{½ }
√(1+ (dy/dx)^{2})
= √(1+ x^{2} / (9-x^{2}))
= √(9 / 9-x^{2})
= 3/ √(9-x^{2})
Here's the integral for the surface area,
S = ∫^{2}_{-2} 2Πy (3/ √(9-x^{2})) dx
Though there is a problem. The meaning of dx here is that we shouldn't have any y's in the integral. Thus, before evaluating the integral we'll require to substitute in for y as well. After that the surface area is,
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