Calculus

Calculus

Postby Guest » Tue Jun 11, 2019 6:10 am

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

Postby Guest » Mon Jun 17, 2019 8:25 am

Imagine dividing the interval from [tex]-\pi[/tex] to [tex]\frac{5}{4}\pi[/tex] into many small intervals of length [tex]dx[/tex]. 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 [tex]\sqrt{1+ f'(x)^2}[/tex] so the surface of that cylinder has length [tex]\sqrt{1+ sin^2(x)}[/tex] so area [tex]\sqrt{1+ sin^2(x)}dx[/tex]. The full area is given by [tex]\int_{-\pi}^{5\pi/4} \sqrt{1+sin^2(x)}dx[/tex].
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