Question

Find the volume of the solid formed by revolving the region bounded by LaTeX: y = \sqrt{x} y = x and the lines LaTeX: y = 1 y = 1 and LaTeX: x = 4 x = 4 about the line LaTeX: y = 1 y = 1 .

Answer 1

The volume is:

`\displaystyle(37\pi)/(10)`

Step-by-step explanation:

See the sketch of the region in the attached graph.

We set the integral using washer method:

`\displaystyle\int_a^b\pi r^2dx`

Notice here the radius of the washer is the difference of the given curves:

`x-√(x)`

So the integral becomes:

`\displaystyle\int_1^4\pi(x-√(x))^2dx`

We solve it:

Factor
`\pi` out and distribute the exponent (you can use FOIL):

`\displaystyle\pi\int_1^4x^2-2x√(x)+x\,dx`

Notice:
`x√(x)=x\cdot x^(1/2)=x^(3/2)`

So the integral becomes:

`\displaystyle\pi\int_1^4x^2-2x^(3/2)+x\,dx`

Then using the basic rule to evaluate the integral:

`\displaystyle\pi\left[(x^3)/(3)-(2x^(5/2))/(5/2)+(x^2)/(2)\right|_1^4`

Simplifying a bit:

`\displaystyle\pi\left[(x^3)/(3)-(4x^(5/2))/(5)+(x^2)/(2)\right|_1^4`

Then plugging the limits of the integral:

`\displaystyle\pi\left[(4^3)/(3)-(4(4)^(5/2))/(5)+(4^2)/(2)-\left((1)/(3)-(4)/(5)+(1)/(2)\right)\right]`

Taking the root (rational exponents):

`\displaystyle\pi\left[(4^3)/(3)-(4(2)^(5))/(5)+(4^2)/(2)-\left((1)/(3)-(4)/(5)+(1)/(2)\right)\right]`

Then doing those arithmetic computations we get:

`\displaystyle(37\pi)/(10)`