`y = e^x, y = sqrt(x) + 1` Use a graph to estimate the x-coordinates of the points of intersection of the given curves. Then use this...

`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...

`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.