- #1
vector22
- 57
- 0
solve this integral
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??
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??
Last edited: