Calculating Volume of Solid using Disk/Washer Method

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SUMMARY

The discussion focuses on calculating the volume of a solid generated by rotating the region enclosed by the curves y=sqrt(x+1), y=2x/3, and the x-axis about the x-axis using the disk/washer method. The user initially attempted to solve the integral V=pi∫(x+1-(4x^2/9))dx over the interval [3,-1], leading to an incorrect volume result. A key insight provided by another participant is to break the problem into two integrals, as the lower boundary changes from y=2/3x to the x-axis within the region of interest.

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Homework Statement


Find the volume of the solid generated by rotating the region enclosed by y=sqrt(x+1), y=2x/3, and the x-axis about the x-axis



Homework Equations


Disk/washer method, shell method



The Attempt at a Solution


This question seemed easy enough, as I found the points of intersection for the 2 functions (these points being -1 and 3), then plugged each of these functions into the disk/washer method equation and tried to solve over [3,-1]. This did not get me very far, for I ended up with an answer higher than the 4pi I should have gained. This is the way I set up the equation:
V=pi\int(x+1-(4x^2/9))dx (I already applied the squares to the functions inside the integral). I am wondering if I made a mistake setting this question up, or if I made a mistake with the closed interval I used. If anyone can point me in the right direction for this question it would be greatly appreciated, thanks in advance.
 
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Hi. You need to plot it carefully and look at it: the lower boundary of the region is the function y=2/3x in one part of the region, but is the x-axis in the second part of the region. So . . . break it up into two integrals.
 

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