Consider the region bounded by the graphs of y=lnx, y=0 and...

Consider the region bounded by the graphs of $y=\ln x, y=0$ and $x=e$ a. Find the area of the region. b. Find the volume of the solid formed by revolving this region about the $x$ -axis. c. Find the volume of the solid formed by revolving this region about the line $x=-2$ d. Find the centroid of the region.

Consider the region bounded by the graphs of $y=\ln x, y=0$ and $x=e$
a. Find the area of the region.
b. Find the volume of the solid formed by revolving this region about the $x$ -axis.
c. Find the volume of the solid formed by revolving this region about the line $x=-2$
d. Find the centroid of the region.

Key Concepts

Volume of Revolution using the Disk/Washer Method

The disk/washer method is used to find the volume of a solid formed by revolving a planar region around an axis. For a typical disk or washer perpendicular to the axis of rotation, one integrates the area of circular cross-sections. This method involves expressing the radius of the disk or washer in terms of the variable of integration and then computing the integral over the appropriate interval.

Volume of Revolution using the Shell Method

The shell method provides an alternative approach for computing the volume of a solid of revolution when the disks or washers are difficult to formulate. It involves slicing the solid into cylindrical shells parallel to the axis of rotation. Each shell has a volume which is the product of its circumference, height, and thickness, and integrating these volumes over the region yields the total volume.

Definite Integration for Area

This concept involves computing the area of a region under or between curves by setting up and evaluating a definite integral. One identifies the correct limits of integration based on the intersection points or specified boundaries and integrates the function representing the top curve minus the bottom curve over this interval.

Centroid of a Planar Region

The centroid (or geometric center) of a region is the point where it would balance if it were made of a uniform material. Determining the centroid involves computing the first moments of the area about the coordinate axes and then dividing by the total area. This requires setting up integrals for the moments and the area, which are then used to find the coordinates of the centroid.

Transcript

00:01 We have different parts of the question and in each of the parts we will be solving the area and volume.

00:11 So part of the problem here is you have to find area which is integration of 1 to e lnx dx.

00:23 So it is written as lnx integration of 1dx minus integration of d upon dx of lnx into integral dx.

00:43 This is written as lnx x into an x into an x minus integration of 1 over x into x d x so it is written as x lennox minus x 1 to e this is written as e lene minus e minus 1 lm 1 minus 1 so it is written as e ln e minus e minus e minus e minus e it is 0, so it is plus over here.

01:23 This becomes a minus a plus 1 which is equal to 1.

01:29 This is the answer to part a of the problem.

01:32 Now we have part b of the problem where we will be finding the volume.

01:45 So part b of the problem is volume is equal to 1 to ax, x, lax.

01:57 So first of all, we'll be finding x.

02:01 So a x is pi, nx, whole square.

02:09 Therefore, volume equals 1 2e, pi, l and x, whole square d x.

02:21 Now this is given as pi 1 2e, l and x, whole square d x.

02:32 This is written as pi into to lnx whole square integration of dx minus diction of dx upon dx of lnx whole square into integral of dx all dx this is written as pi into x lnx whole square minus integration of 2 lnx into one over of x x d x is written as pi x telenex whole square minus twice integration of telenx d x is written as pi into x x tell n x whole square minus 2 into x and the limits are 1 and 2.

03:58 Now we have this further written as pi into e l &e whole square minus 2 into e lne minus e minus pi 1 1 whole square minus 2 into 1 so this is 0 over here.

04:38 This becomes pi into e minus 2 into...

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