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Find the Volume: y=sqrt(x+2), y=x, y=0, revolved around the x-axis using the shell method
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@dumbcow if you could help please! :)
ok sorry about wait, this one was tripping me up for a bit sometimes i would rather use washer method anyway looks like you need to split it into 2 separate integrals find volume of parabola part, then subtract volume of triangular part |dw:1380596700713:dw| \[V = 2\pi \int\limits_0^2 y(4-y^{2}) dy - 2\pi \int\limits_0^2 y(2-y) dy\]
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