Help! Find the Volume of the solid!

carlycakes

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Find the Volume of the solid generated by revolving the region about the y-axis.

the region enclosed by x = y^2 / 5 , x = 0, y = -5, y = 5

can you please show me step-by-step process to get the answer! thanks! :p
 


Please tell me what you've learned so far about the Disc Method for finding volumes of solids of revolution.

It would also help, if you could say something about what you've already done OR where you're stuck.

Thanks 8-)

PS: This appears to be two identical objects, not one, since there is no volume at the Origin. We can integrate from y = 0 to y = 5, and double the result.
 
im stuck on the problem...the disk method is soo confusing.
i don't know where to start.
 
Hello, carlycakes!

Find the Volume of the solid generated by revolving the region about the y-axis.

. . x=y25,  x=0,  y=5,  y=5\displaystyle x = \frac{y^2}{5},\;x = 0,\; y = -5,\; y = 5

First, make a sketch . . .


Code:
       5|
        |:::::::::*
        |:::::*
        |::*
        |*
    - - * - - - - - -
        |*
        |::*
        |:::::*
        |:::::::::*
      -5|

How can "discs" be confusing?
It's the simplest of all the Volume-of-Revolution problems . . .

Formula:   V  =  π ⁣ ⁣ab ⁣ ⁣x2dy\displaystyle \text{Formula: }\;V \;=\;\pi\!\! \int^b_a\!\! x^2\,dy


We have:   V    =    π ⁣-55 ⁣(y25)2dy    =    π25 ⁣-55 ⁣y4dy    =    π125y5]-55\displaystyle \text{We have: }\;V \;\;=\;\;\pi\!\int^5_{\text{-}5}\!\left(\frac{y^2}{5}\right)^2dy \;\;=\;\;\frac{\pi}{25}\!\int^5_{\text{-}5}\!y^4\,dy \;\;=\;\; \frac{\pi}{125}y^5\,\bigg]^5_{\text{-}5}

. . . . . . . . =    π125[(55)(5)5]    =    π53(255)    =    π252    =    50π\displaystyle =\;\;\frac{\pi}{125}\bigg[(5^5) - (-5)^5\bigg] \;\;=\;\; \frac{\pi}{5^3}\left(2\cdot5^5\right) \;\;=\;\;\pi\cdot2\cdot5^2 \;\;=\;\;50\pi

 
Given: x = y25, y = ±5, x = 0, find volume by rotating along the yaxis.\displaystyle Given: \ x \ = \ \frac{y^2}{5}, \ y \ = \ \pm5, \ x \ = \ 0, \ find \ volume \ by \ rotating \ along \ the \ y-axis.

Method 1, disks: 2π05(y25)2dy = 50π.\displaystyle Method \ 1, \ disks: \ 2\pi\int_{0}^{5}\bigg(\frac{y^2}{5}\bigg)^2dy \ = \ 50\pi.

Method 2, shell: 4π05x(55x)dx = 50π.\displaystyle Method \ 2, \ shell: \ 4\pi\int_{0}^{5}x(5-\sqrt{5x})dx \ = \ 50\pi.

See graph.\displaystyle See \ graph.

[attachment=0:2e2gfc6y]eee.jpg[/attachment:2e2gfc6y]
 

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