FInd the volume of the solid formed by revolving the region bounded by f(x)= (sq

Question

FInd the volume of the solid formed by revolving the region bounded by f(x)= (squareroot x) + 2, x=0, and x=4, about the line y=2.

Explanation / Answer

f(x)= top or left most function within limits g(x)= bottom or right most function within limits 1. Set functions equal to each other for limits of integration (bounds). Use formula -> int from a to b of 2*pi*x*( f (x) - g (x) ) dx 2. Set functions equal to each other for limits of integration. Then Int from a to b of f(x)-g(x). 3. Use the shell formula with x=1 and x=0 as the limits of integration and y= 4e whatever as f(x) and y=0 as g(x) 4. Find the zero's of the function, those will be your limits of integration. integrate with the shell formula like in number three with the given function as f(x) and y=0 as g(x). You can also use the washer formula and it should give the same answer: int from a to b of pi*( f^2 - g^2 ). 5. Volume of a sphere w/ radius 4 - Volume of a cylinder with radius 1. 6. Use Disk formula: int from a to b of pi* f^2. x=0 and x=7 are limits of integration. 7. set the two y=... equations equal to each other. This is your left limit. x=2 is the right limit. Use washer method integration where f is y=4 and g is y=4-x^2. 8. Washer method: the two x=... equations are your limits and the functions are y=... 9. Shell method: set the y=... equations equal to each other for left bound and use x=6 for right bound. y=2 is f and y=6/x is g. 10. Use x=0 for left bound and set y=... equations equal to each other for right bound. y=6 is f and y=6sqrt(x) is g. Use either shell or washer formula to calculate. (I prefer shell).

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