Use the result of Exercise 37 in Section 6.1 (or the exercise above if your answer is correct) to find the volume V of the solid obtained by revolving about the line y = - 5 the region between the graph of the equation y = sqrt(x + 1) on the interval [0, 1].
V =∫ =
assuming you mean the region between the graph and the x-axis, that gives
v = ∫[0,1] π(R^2-r^2) dx
where R = y+5 and r = 5
v = ∫[0,1] π((√(x+1)+5)^2-5^2) dx
π/2[(1+1)^(3/2) - (0+1)^(3/2)]
0 (sqrt(x+1))^2 dx
V = π/2[2^(3/2) - 1]
V = π/2[4.89898 - 1]
V = π/2(3.89898)
V = 5.94897π