Having a little bit of struggle with this cal2 question.

[half ore]

the answer was given with the paper but Im not sure what the method is to get it

[still perch]

Note that if you move everything down 2 units, you have the region bounded by y = x^2 - 2 and y = -1 rotated about the y-axis, but it's exactly the same shape because you moved everything in the same direction.

[half ore]

ok....... and?

[still perch]

...what is the actual difficulty you have?

[plain sandal]

washer method would get π∫(x²-2)²dx after you evaluate your region of integration (x²-2=-1 factors to (x-1)(x+1) so your limits are -1→1)
from here you have the solid you want and a disc within it
the disc is 2 thick and of radius 1, so you now have
π∫(x²-2)²dx-2π

@half ore

[half ore]

thank u