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Determining the Surface Area of a Solid of Revolution

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How do I find the surface area of the solid defined by revolving $r = 3 sin (θ)$ $r = 3sin(theta)$ about the polar axis?

Write :

$r = 3 sin (θ)$ $r = 3 sin(theta)$

$r = 3 \frac{y}{r}$ $r = 3y/r$ because $y = r sin (t)$ $y=r sin(t)$

$r^{2} = 3 y$ $r^2 = 3y$

$x^{2} + y^{2} = 3 y$ $x^2 + y^2 = 3y$ because $r = \sqrt{x^{2} + y^{2}}$ $r = sqrt(x^2+y^2)$ .

$x^{2} + {(y - \frac{3}{2})}^{2} = \frac{9}{4}$ $x^2 + (y-3/2)^2 = 9/4$

You recognize a circle of radius $\frac{3}{2}$ $3/2$ . The area is $π {(\frac{3}{2})}^{2} = 9 \frac{π}{4}$ $pi (3/2)^2 = 9 pi/ 4$ .

vince · 1 · Feb 24 2015
How do I find the surface area of a solid of revolution using polar coordinates?

If a surface is obtained by rotating about the x-axis the polar curve $r = r (θ)$ $r=r(theta)$ from $θ = θ_{1}$ $theta=theta_1$ to $θ_{2}$ $theta_2$ , then its surface area A can by found by

$A=2pi int_{theta_1}^{theta_2}r(theta)sin theta sqrt{r^2+[r'(theta)]^2} d theta$ .

I hope that this was helpful.

Wataru · · Oct 18 2014

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Polar Curves

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