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Nivelo

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Question: An area in the first quadrant (x=>0,y=>0) is limited by the axis and the graphs to the functions f(x)=x^2-2 and g(x)=2+x^2/4. When the area rotates around the y-axis a solid is created. Calculate the volume of this solid.

I want to calculate with the disc method. I set h(x)=g(x)-f(x) since g(x)>f(x) in this area. Since x>=0 i check where the lower function crosses the x-axis and get

x^2-2=0 which gives x=sqrt(2) (since x can't be negative). The upper limit is where f(x)=g(x), so

x^2-2=x^2/4+2 which gives x=4/sqrt(3).

One of the discs that i want to sum up should have the volume pi*radius^2*dy.I get the radius from solving x^2 from h(x)=y.And so i get that the volume should be

I=pi/3* integral from 0 to 10/3 of (16-4*y)dy. But its not the right answer and i don't know where i went wrong. Could anyone tell me where i messed up and try to help me solve this one?

Thanks in advance, appreciate all the help!