Question

The function f and g are given by
`f(x)=x^2` and
`g(x)=-^1_2 x+5`.

Let R be the region bounded by the x-axis and the graphs f and g as shown at the bottom. Also, I only need you to answer part b, I've already finished part a.


a) Describe how you would find the area of R.



`R=18 (2)/(3)`


b) The region R is the base of a solid. For each y, where
0 < y < 4, the cross-section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 3y. Explain how you would write an expression that gives the volume of the solid.

Answer 1

As you've pointed out, R is a different region that I originally thought. The area of R can be computed using either

`\displaystyle \int_0^2 x^2 \, dx + \int_2^(10) -\frac x2 + 5 \, dx`

or

`\displaystyle \int_0^4 (10-2y)-\sqrt y \, dy`

Either way, you've gotten the correct area for part (a).

For part (b), we can use part of the integral with respect to
`y` above. The horizontal distance between the curves
`y=x^2` and
`y=-\frac x2+5` is obtained by first solving for
`x`,

`y=x^2 \implies x=\sqrt y`

`y=-\frac x2+5 \implies x = 10-2y`

so the length of each cross section is
`10-2y-\sqrt y`.

The height of each cross section is
`3y`.

Then the volume of the solid is

`\displaystyle \int_0^4 3y \left(10-2y-\sqrt y\right) \, dy = \boxed{\int_0^4 -6y^2 + 30y - 3y^(3/2) \, dy}`

The instructions don't say to evaluate, but if you're looking for practice the volume ends up being 368/5, or 73 3/5.