The area the surface generated by revolving the curve about the given axis stated above is: 98π
Because runs from 0 to pi/2 and rotation occurs about the y axis, surface area is given by:
[tex]\small S = \int _0^{\frac{\pi}{2}}2\pi xds[/tex] ----------------------------------------------------1)
This is so where;
ds = [tex]\sqrt{[(dx/dt)^{2}} + (dy/dt)^{2}] dt[/tex]
ds = [tex]\sqrt{([dx/d\theta)^{2}}+(dy/d\theta)^{2}] d\theta[/tex]
ds = [tex]\sqrt{([dx/d\theta (7cos\theta))^{2}}+(dy/d\theta (7cos\theta))^{2} d\theta][/tex]
ds = [tex]\sqrt{[(-7cos\theta)^{2}} + (-7cos\theta)^{2}] \theta[/tex]
ds = [tex]\small \sqrt{49sin^2(\theta)+49cos^2(\theta) }d\theta[/tex]
ds = [tex]\small \sqrt{49(sin^2(\theta)+cos^2(\theta)) }d\theta[/tex]
ds = √49dθ
ds = 7dθ
Hence, we can say that,
[tex]\small S = \int _0^{\frac{\pi}{2}}2\pi (7cos(\theta))(7d\theta)[/tex]
[tex]\small S =98\pi \int _0^{\frac{\pi}{2}}cos(\theta)d\theta[/tex]
[tex]\small S =98\pi \left [ 1-0 \right ][/tex]
S = 98π
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Full Question:
Find the area of the surface generated by revolving the curve about the given axis. x = 7 cos(θ), y = 7 sin(θ), 0 ≤ θ ≤ π 2 , y-axis
