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.
Duplicate post. See my previous answer.
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 to get a better understanding of the problem.
The region enclosed by the curves is a bounded region between the curve y = e^(x+2) and the x-axis, from x = 0 to x = 0.7. The region looks like a curve that starts from the point (0,1) and asymptotically approaches the x-axis as x increases.
Next, we'll express the given curve equation in terms of y and rewrite it in terms of x. Let's solve for x in terms of y:
y = e^(x+2)
Take the natural logarithm (ln) of both sides:
ln(y) = x + 2
Rearrange the equation to solve for x:
x = ln(y) - 2
Now that we have x in terms of y, let's find the volume of the solid by integrating the function 2πxy with respect to x.
The volume V can be expressed as follows:
V = ∫[a,b] 2πxy dx
Since we need to rotate the region about the x-axis from x = 0 to x = 0.7, our definite integral limits will be from x = 0 to x = 0.7:
V = ∫[0,0.7] 2π(xy) dx
Substituting x from our expression in terms of y:
V = ∫[0,0.7] 2π((ln(y) - 2)y) dx
Now, we need to replace dx with its equivalent in terms of y. Since x is changing, dx becomes dy. We'll also change the limits of the integral from x = 0 to x = 0.7 to y = 0 to y = e^(0+2) = e^2:
V = ∫[0,e^2] 2π((ln(y) - 2)y) dy
Integrating this expression will give us the volume of the solid formed by rotating the region enclosed by the given curves about the x-axis.