# Calculus

### Calculus

The Area Bounded by the Curve y=sin(x), x = -π, x = $$\frac{5}{4}$$π is rotated about the x-axis. Find The Area of Surface of Revolution?
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### Re: Calculus

Imagine dividing the interval from $$-\pi$$ to $$\frac{5}{4}\pi$$ into many small intervals of length $$dx$$. Rotating around the x-axis each small section rotates to a short cylinder of radius y= |sin(x)|. A short section of y= f(x) will have length $$\sqrt{1+ f'(x)^2}$$ so the surface of that cylinder has length $$\sqrt{1+ sin^2(x)}$$ so area $$\sqrt{1+ sin^2(x)}dx$$. The full area is given by $$\int_{-\pi}^{5\pi/4} \sqrt{1+sin^2(x)}dx$$.
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