Calculus: Early Transcendentals, Chapter 6, 6.3, Section 6.3, Problem 33

y=e^x
y=sqrtx + 1
The graph of these two equations are:

(Green curve graph of y=e^x . And blue curve is the graph of y=sqrt(x) + 1 .)
Base on the graph, the two curve intersect at x=0 and x~~0.56 .
To solve for the volume of the solid formed when the bounded region is rotated about the y-axis, apply the method of cylinder. Its formula is:
V=int_a^b 2pi*r*h*dx
To determine the radius and height of the cylindrical shell, refer to the figure below. Its radius and height are:
r = x
h=y_(upper) - y_(lower)=sqrtx + 1 - e^x
Plug-in them to the formula of volume.
V=int_0^0.56 2pi *x*(sqrtx + 1-e^x) dx
V= 2pi int _0^0.56 (x^3/2 + x - xe^x) dx
Take the integral of each term.
V= 2pi (int_0^0.56dx + int_0^0.56 xdx - int_0^0.56 xe^xdx)
For the first two integral, apply the formula int x^n dx = x^(n+10)/(n+1) .
And for the third integral, apply integration by part int u dv = uv - int vdu .
V=2pi( (2x^(5/2))/5+x^2/2 - (xe^x-e^x))|_0^0.56
V = 2pi ( (2x^(5/2))/5+x^2/2 - xe^x+e^x)|_0^0.56
V = 2pi [( (2*0.56^(5/2))/5+0.56^2/2-0.56*e^0.56+e^0.56)- ((2*0^(5/2))/2+0^2/2-0*e^0+e^0)]
V=0.1317
Therefore, the volume of the solid formed is 0.1317 cubic units.