How do you find the volume of the solid generated by revolving the region bounded by #y= 2x-1# #y= sqrt(x)# and #x=0# and revolve about the y-axis?

Question

Christopher Agresta · Accepted Answer

To avoid doing 2 integral (which the disc method would require), I would use cylindrical shells.

A representative slice parallel to the #y# axis has area #(sqrtx - (2x-1))dx#.

Revolving around the #y# axis, gives us a shell of volume:

#2 pi x (sqrtx - 2x+1))dx#. The volume of the solid is:

#2 pi int_0^1 x (sqrtx - 2x+1)dx = 2 pi int_0^1 (x^(3/2) - 2x^2+x)dx #

That's a straightforward integral to evaluate and finish the problem.

You should get #(7pi)/15#.

Alternative:

You could use discs (washers), but you'l have to #int_-1^0# using the

difference between the lines #x=0# and #x=(y+1)/2# And then

#int_0^1# using the difference between #x=(y+1)/2# and #x=y^2#.

Luke Alvarez · Answer

undefined To find the volume of the solid generated by revolving the region bounded by $ y = 2x - 1 $, $ y = \sqrt{x} $, and $ x = 0 $ about the y-axis, you can use the method of cylindrical shells. The volume $ V $ is given by the formula:

$$ V = 2\pi \int_{a}^{b} x \cdot (f(x) - g(x)) \, dx $$

where $ f(x) $ and $ g(x) $ are the upper and lower functions respectively, and $ a $ and $ b $ are the x-values where they intersect.

First, find the intersection points of the curves:

$$ 2x - 1 = \sqrt{x} $$

$$ (2x - 1)^2 = x $$

$$ 4x^2 - 4x + 1 = x $$

$$ 4x^2 - 5x + 1 = 0 $$

$$ (4x - 1)(x - 1) = 0 $$

$$ x = \frac{1}{4}, \, x = 1 $$

So, $ a = \frac{1}{4} $ and $ b = 1 $.

Now, subtract the lower function from the upper function and integrate:

$$ V = 2\pi \int_{\frac{1}{4}}^{1} x \cdot ((2x - 1) - \sqrt{x}) \, dx $$

$$ V = 2\pi \int_{\frac{1}{4}}^{1} (2x^2 - x - x\sqrt{x}) \, dx $$

$$ V = 2\pi \left( \frac{2}{3}x^3 - \frac{1}{2}x^2 - \frac{2}{5}x^{\frac{5}{2}} \right) \bigg|_{\frac{1}{4}}^{1} $$

$$ V = 2\pi \left( \frac{2}{3} - \frac{1}{2} - \frac{2}{5} - \frac{1}{192} + \frac{1}{32} + \frac{1}{10} \right) $$

$$ V = \frac{7\pi}{15} $$

So, the volume of the solid generated by revolving the region about the y-axis is $ \frac{7\pi}{15} $.