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

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?

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

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

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

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