Asked Feb 16, 2019

Use the best method available to find the volume of the region bounded by y=e^x-1, y=2-x and the x-axis revolved about the (a) x-axis and (b) y-axis.


Expert Answer

Step 1

Question has two sub parts. Each of them is quite lengthy. I have run out of steps that can be added in the answer board, while solving the second part. Hence, I will solve the first part completely in detail. Please post the second part as a separate question.

Please see the picture attached on the white board. The curve y = ex-1 and y = 2 - x are plotted. They intersect at point P. The shaded portion is the area enclosed that is first rotated across x - axis.

Let's divide the volume so generated in to two parts. Part A that is bounded by the curve y = ex-1 and part B bounded by the straight line y = 2 - x.

A slice in region A is shown in the second picture. It's radius R = y = ex-1.

Area, A  = pi.R2 = pi.( ex-1)2; Hence, dV = A.dx = pi.( ex-1)2.dx

A slice in region B is shown in green ink. It's radius, R = y = 2 - x

Area, A  = pi.R2 = pi.(2 - x)2; Hence, dV = A.dx = pi.(2 - x)2.dx

Step 2

Now we are ready to integrate the slices in two different regions and add the two to get the desired volume. However, there is a problem, we don't know where exactly point P is located. Point P is point of interesection of the two curves. We will have to solve for this point. At this point, y value from both the curves is same.

Hence, ex-1 = 2 - x; Hence, x + ex = 3

We will not be able to solve this manually. Let's try hit and trial. At x =1, LHS = ex + x = 3.71 > RHS. Hence the point of intersection has to be before x = 1. Try x = 0.9 and then 0.8 and then 0.79. At x = 0.792; LHS = 2.9999 which is nearly equal to 3.

Step 3

Hence, region A needs to be integrated over x = 0 to x = 0.792 and regiob B needs to be integrated over x = 0.792 to x = 2.

Let's call the volume of ...

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