Find the volume of the solid formed by rotating the region enclosed by
y=e^(1x+2), y=0 ,x=0 , x=0.7
about the x-axis.
Why do you have a 1 in front of the x?
I suspect you have copied the problem incorrectly.
Anyway, to get the answer, compute
Integral of pi y^2 dx
x=0 to 0.7
To find the volume of the solid formed by rotating the region enclosed by the given curves about the x-axis, we can use the method of cylindrical shells.
First, let's sketch the region enclosed by the curves y = e^(x+2), y = 0, x = 0, and x = 0.7 on a graph.
The region is a trapezoidal shape between the x-axis and the curve y = e^(x+2) for x ranging from 0 to 0.7. We want to rotate this region about the x-axis to create a solid.
To find the volume of this solid using cylindrical shells, we need to integrate the area of each cylindrical shell over the range of x.
The volume of each cylindrical shell is given by:
dV = 2πrhdx
Where r is the radius of each cylindrical shell, h is the height of each cylindrical shell, and dx is the thickness of each cylindrical shell.
To find the radius r of each cylindrical shell, we can use the distance between the x-axis and the curve y = e^(x+2). So, r = e^(x+2).
To find the height h of each cylindrical shell, we can use the difference between the upper and lower y-values of the region. So, h = e^(x+2) - 0 = e^(x+2).
Now, we can rewrite the volume equation as:
dV = 2π(e^(x+2))(e^(x+2))dx
To find the total volume, we need to integrate this equation from x = 0 to x = 0.7:
V = ∫(0.7 to 0) 2π(e^(x+2))(e^(x+2))dx
Solving this integral will give us the volume of the solid formed by rotating the region about the x-axis.