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

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.


[tex]R=18 \frac{2}{3}[/tex]


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.

The function f and g are given by texfxx2tex and texgx12 x5texLet R be the region bounded by the xaxis and the graphs f and g as shown at the bottom Also I only class=

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LammettHash

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

[tex]\displaystyle \int_0^2 x^2 \, dx + \int_2^{10} -\frac x2 + 5 \, dx[/tex]

or

[tex]\displaystyle \int_0^4 (10-2y)-\sqrt y \, dy[/tex]

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

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

[tex]y=x^2 \implies x=\sqrt y[/tex]

[tex]y=-\dfrac x2+5 \implies x = 10-2y[/tex]

so the length of each cross section is [tex]10-2y-\sqrt y[/tex].

The height of each cross section is [tex]3y[/tex].

Then the volume of the solid is

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

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.

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