This week we learned the disc method and the washer method. Grasping the concepts of these methods is pretty difficult. It's hard to picture rotating a solid around an axis in your head. The disc method is used for solids. When rotating a function around an axis, I'd first graph the function, then draw in a representative rectangle, then label the rectangle accordingly. Next I would write my integral. For discs, the volume would be v=pi * the integral from a to b of f(x)^2 dx. Then I'd plug that into NINT on my calculator to get the final answer.
An example of a disc problem would be f(x)=(sinx)^.5 , from 0 to pi , around the x-axis. The graph of this function looks kind of like a semi-circle from 0 to pi. The representative rectangle would be vertical because the integral is in terms of x. the width of the rectangle would be dx while the heigh would be f(x) which was given f(x)=(sinx)^.5. My integral would be pi * the integral from 0 to pi of ((sinx)^.5)^2 dx. After integrating this, my final answer would be 2 pi.
Washers are basically discs with a cylindrical hole in the middle. The volume for a washer would be V=pi * the integral from a to b of R(x)^2 - r(x)^2 dx. An example of a washer problem would be y=x^.5 , y=x^2 , around the x-axis. The representative rectangle would be vertical because the integral will be in terms of x. The width of the rectangle is dx, the big radius of the function is x^.5, and the small radius of the function is x^2. The integral would be V=pi * the integral from 0 to 1 of ((x^.5)^2) - (x^2)^2 dx. My final answer would be 3pi/10.
An example of a disc problem would be f(x)=(sinx)^.5 , from 0 to pi , around the x-axis. The graph of this function looks kind of like a semi-circle from 0 to pi. The representative rectangle would be vertical because the integral is in terms of x. the width of the rectangle would be dx while the heigh would be f(x) which was given f(x)=(sinx)^.5. My integral would be pi * the integral from 0 to pi of ((sinx)^.5)^2 dx. After integrating this, my final answer would be 2 pi.
Washers are basically discs with a cylindrical hole in the middle. The volume for a washer would be V=pi * the integral from a to b of R(x)^2 - r(x)^2 dx. An example of a washer problem would be y=x^.5 , y=x^2 , around the x-axis. The representative rectangle would be vertical because the integral will be in terms of x. The width of the rectangle is dx, the big radius of the function is x^.5, and the small radius of the function is x^2. The integral would be V=pi * the integral from 0 to 1 of ((x^.5)^2) - (x^2)^2 dx. My final answer would be 3pi/10.