Let V be the volume of the solid obtained by rotating about the y-axis the region bounded by y=√x and y=x^2. Find V either by slicing (the disk/ washer method) or by cylindrical shells.

Since the rotation is around the y-axis we need to change de dependency of our variables to have [tex]f(x)\rightarrow f(y)[/tex]. Thus, our functions with [tex]y[/tex] as independent variable are:

[tex]x=\sqrt{y}\\ x=y^2[/tex]

For the washer method, we need to find the area function, which is given by:

[tex]A=\pi\cdot [(\rm{outer\ radius)^2 -(\rm{inner\ radius)^2 ][/tex]

By taking a look at the plot I attached, one can easily see that for a rotation around the y-axis the outer radius is given by the function [tex]x=\sqrt{y}[/tex] and the inner one by [tex]x=y^2[/tex]. Thus, the area function is:

[tex]A(y)=\pi\cdot [(\sqrt{y} )^2-(y^2)^2]\\A(y)=\pi\cdot (y-y^4)[/tex]

Now we just need to integrate. The integration limits are easy to find by just solving the equation [tex]\sqrt(y)=y^2[/tex], which has two solutions [tex]y=0[/tex] and [tex]y=1[/tex]. These are then, our integration limits.

[tex]V=\pi\int_{0}^1 (y-y^4)dy\\ V=\pi (\int_{0}^1 ydy - \int_{0}^1 y^4dy)\\ V=\pi/2-\pi/5\\\boxed{V=3\pi/10}[/tex]

xero099 xero099 · Answer 2 · 2022-02-17T13:18:01+01:00

The volume of the solid obtained by rotating about the y-axis the region is [tex]\frac{9}{70}\pi[/tex] cubic units.

How to determine the volume of a solid of revolution generated by a region comprised by two functions

The volume of a solid of revolution for a region comprised by two functions and that rotates about the x-axis is described by following integral formula:

[tex]V = \pi\int\limits^1_0 {[f(y)-g(y)]^{2}} \, dy[/tex]

Please notice that points of intersection are [tex](x_{1},y_{1}) = (1,1)[/tex] and [tex](x_{2},y_{2}) = (0,0)[/tex].

If we know that [tex]f(y) = y^{2}[/tex] and [tex]g(y) = y^{0.5}[/tex], then the integral formula is:

[tex]V = \pi\int\limits^1_0 {[y^{2}-y^{0.5}]^{2}} \, dy[/tex]

[tex]V = \pi\int\limits^{1}_{0} {y^{4}-2\cdot y^{2.5}+y} \, dy[/tex]

[tex]V = \pi\left[\int\limits^1_0 {y^{4}} \, dy -2\int\limits^1_0 {y^{2.5}} \, dy +\int\limits^1_0 {y} \, dy\right][/tex]

[tex]V=\pi\cdot \left[\frac{y^{5}}{5}|_{0}^{1}-2\cdot \left(\frac{y^{3.5}}{3.5} \right)|_{0}^{1} + \frac{y^{2}}{2}|_{0}^{1}\right][/tex]

[tex]V = \frac{9}{70} \pi[/tex]

The volume of the solid obtained by rotating about the y-axis the region is [tex]\frac{9}{70}\pi[/tex] cubic units. [tex]\blacksquare[/tex]

To learn more on solids of revolution, we kindly invite to check this verified question: https://brainly.com/question/338504