I know it is
pi (integral from zero to 1)then I get confused
Which of the following integrals will find the volume of the solid that is formed when the region bounded by the graphs of y = ex, x = 1, and y = 1 is revolved around the line y = -2
2 answers
I assume you have drawn the region. It's a triangular shape with vertices at (0,1), (1,1) and (1,e).
So, using discs (washers) of thickness dx,
v = ∫[0,1] π(R^2-r^2) dx
where R=y+2 and r=3
v = ∫[0,1] π((e^x+2)^2-3^2) dx
= π/2 (e^2+8e-19)
Or, using shells of thickness dy,
v = ∫[1,e] 2πrh dy
where r=y+2 and h=1-x
v = ∫[1,e] 2π(y+2)(1-ln y) dy
= π/2 (e^2+8e-19)
So, using discs (washers) of thickness dx,
v = ∫[0,1] π(R^2-r^2) dx
where R=y+2 and r=3
v = ∫[0,1] π((e^x+2)^2-3^2) dx
= π/2 (e^2+8e-19)
Or, using shells of thickness dy,
v = ∫[1,e] 2πrh dy
where r=y+2 and h=1-x
v = ∫[1,e] 2π(y+2)(1-ln y) dy
= π/2 (e^2+8e-19)
1 answer