1
$\begingroup$

Find the volume $V_y$ of the figure bounded by the lines $y=x^2,x=3,y=0$

I tried:

$$V_y=\pi \int_{0}^{3} x^4 dx=\pi \frac {3^5}{5}$$

Is this solution correct?

$\endgroup$
4
  • 2
    $\begingroup$ Are you referring to a revolution solid around x axis? $\endgroup$
    – user
    May 3 '18 at 9:49
  • $\begingroup$ Around $O_y$ .. $\endgroup$
    – user548054
    May 3 '18 at 9:50
  • $\begingroup$ You should precise this point in the OP, are you looking for the volume of the solid of revolution around the y axis? $\endgroup$
    – user
    May 3 '18 at 9:52
  • $\begingroup$ Yes you are right.. $\endgroup$
    – user548054
    May 3 '18 at 9:53
1
$\begingroup$

Your derivation is correct for the volume for the solid of revolution around the $x$ axis by disk method.

Note that, for to the volume for the solid of revolution around the $y$ axis by disk method the set up would be

$$\pi \int_{0}^{9} (9-y)\, dy$$

$\endgroup$
5
  • $\begingroup$ How did you find this formula? $\endgroup$
    – user548054
    May 3 '18 at 9:57
  • $\begingroup$ Opsss there was a typo. To see that make a sketch; for each fixed value of y the elementary volume is $dV=(\pi R_2^2-\pi R_1^2)dy=\pi (3^2-(\sqrt y)^2)dy$ $\endgroup$
    – user
    May 3 '18 at 10:02
  • $\begingroup$ @Beginner You can also refer to tutorial.math.lamar.edu/Classes/CalcI/VolumeWithRings.aspx $\endgroup$
    – user
    May 3 '18 at 10:04
  • $\begingroup$ Thank you very much. (+1) $\endgroup$
    – user548054
    May 3 '18 at 10:18
  • 1
    $\begingroup$ @Beginner You are welcome! Bye $\endgroup$
    – user
    May 3 '18 at 10:19