Find the area using disk method

Creating a new topic here requires registration. All other forums are without registration.

Find the area using disk method

Postby jaychay » Thu Oct 22, 2020 8:32 am

If [tex]V_1 - V_2 = 4\pi[/tex] find R
Attachments
Untitled 1.png
Untitled 1.png (11.89 KiB) Viewed 878 times
jaychay
 
Posts: 5
Joined: Tue Sep 01, 2020 6:58 pm
Reputation: 0

Re: Find the area using disk method

Postby Guest » Sun Dec 19, 2021 12:50 pm

It would be nice if they told us what "[tex]V_1[/tex]" and "[tex]V_2[/tex]" stood for!

I am going to guess that [tex]V_1[/tex] is the volume of the entire region from the x-axis up to the graph of y= f(x) between x= 1 and x= 4, rotated around the x-axis, [tex]--\pi \int_1^4 f^2(x)dx[/tex], and that [tex]V_2[/tex] is the volume of the region from the x-axis up to y= 1 rotated around the x-axis. That last is just a cylinder of radius 1 and length 3 so its volume is [tex]3\pi[/tex].

That is, [tex]\pi \int_1^4 f^2(x)dx- 3\pi= 4\pi[/tex] so you know that [tex]\pi \int_1^4 f^2(x)dx= 7\pi[/tex] and you are asked to find [tex]\int_1^4 f(x)dx[/tex] minus the area below the shaded region which is 4.

(It occurs to me that we could use the "theorem of pappus" which says that the volume of a region, rotated around the x-axis, is the area of the region times the distance from the x-axis to the centroid of the area, but that requires finding the centoid!)
Guest
 


Return to Math Problem of the Week



Who is online

Users browsing this forum: No registered users and 1 guest