Question

Use the shell method to find the volume about the x-axis? y=1/2x^2 and y=0 around the x-axis from x=0 to x=6

Answer 1

the answer
`v_(shell)=int_0^18y*[6-(2y)^(1/2)]*dy=[(60*y^2-2^(7/2)*y^(5/2))/20]_0^18=194.4`

Explanation:

In the shell when the curve revolve around the x-axis the slice should be parallel to the x-axis

so we will find the value of y

when x=0 the value of y equal y=0

when x=6 the value of y equal y=18

`y=1/2*x^2`

`x=sqrt(2y)`

the shaded region is revolving

now will set up our integral

`v_(shell)=int_0^18y*[6-sqrt(2y)]*dy`

`v_(shell)=int_0^18y*[6-(2y)^(1/2)]*dy=[(60*y^2-2^(7/2)*y^(5/2))/20]_0^18=194.4`