How to find the volume of the solid generated with the given information

calculusdefinite integralsintegration

Here is the question:

Find the volume of the solid generated by revolving the regions bounded by the graphs of the equation about the y-axis.

Given: y=$\sqrt{x}$; $y=0$; $x=3$

Here are the steps I took to get the (wrong) answer:

  1. Area = $\pi \times (y^2)^2 = \pi \times y^4$
  2. $\int_{0}^{\sqrt{3}}(\pi y^4) dy$
  3. After evaluating this definite integral, I got $9{\pi} \sqrt{3}$ all over 5.

Can you please help me figure out where I made my mistake or if I took the wrong approach?
Thank you.

Best Answer

This region:

Mathematica graphics

It is one of two congruent finite area regions that have edges on all three bounding curves. (The other one is mirror symmetric below the $x$-axis and has the same volume of revolution.)

This is a horizontally simple region. You should be able to find the volume of revolution (about the $y$-axis) using $$ \pi \int_0^\sqrt{3} (3)^2 - (\text{something})^2 \,\mathrm{d}y \text{,} $$

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