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The area of a circle can be found by the washer method

The exact area of a washer is

[tex] dA = 2 \pi r \,\,dr \,\,\,\,\,\,\,eq.1[/tex]

Area of a circle is:

[tex] \int 2 \pi r \,\,dr [/tex]

Equation 1 mulitplied by h (height) gives the exact volume of a cylindrical shell:

[tex] dV = 2 \pi r h\,\,dr [/tex]

From here it is possible to use trig functions to calculate the volume of a hemisphere by the method of cylindrical shells.

so

[tex] r = cos \theta [/tex]

[tex] h = sin \theta [/tex]

[tex] dr = -sin \theta \,\, d\theta[/tex]

putting it all together

[tex] \int -sin^2 \theta \, cos \theta \,\, d\theta [/tex]

Anyway, how do you solve that last integral??

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# Surface area and volume of a sphere

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