Find Volume of Rotated Region Bounded by y=x & y=[tex]\sqrt{x}[\tex]

[merced]

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.

y = x
y = [tex]\sqrt{x}[\tex]<br /> rotate about y = 1<br /> <br /> <a href="http://imageshack.us" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://img461.imageshack.us/img461/5879/math10sp.jpg </a><br /> <br /> =<a href="http://img161.imageshack.us/my.php?image=math23gk.jpg" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://img161.imageshack.us/img161/5729/math23gk.th.jpg </a><br /> <br /> So, I am integrating with respect to x.<br /> Area = [tex]\int^1_{0}\pi[(f(x))^2-(g(x))^2]dx[\tex]<br /> <br /> I assumed that f(x) = x and g(x) = [tex]\sqrt{x}[\tex].<br /> <br /> However, the book gives f(x) = 1 - x and g(x) = 1 - $$\sqrt{x}[\tex].<br /> <br /> I don't understand how they got that.$$[/tex][/tex][/tex]

 

[Curious3141]

Double post. Already replied here : https://www.physicsforums.com/showthread.php?t=121106

 

[Lyuokdea]

what axis are you rotating the function around and how does that affect the volumes you will get?

A nicer working problem to go through, which might better illustrate the principals is this: Find the volume generated by rotating the area in between the two functions: y=1 and y=1/2 around the axis y=1. How do the two lines function when they are rotated?

Note: You will be able to check your answer to this problem by comparing the answer with the volume of a cylinder.

Curious: beat me to it I guess on the other thread.

~Lyuokdea