Consider the functions f(x)=x^3/(x^4+1) and g(x)=x^2/(x^4+1). Let R denote the region in the first quadrant bounded by the curves y=f(x) and y=g(x).
Find the exact volume of the solid that has R as its base if every cross section by a plane perpendicular to the x-axis is a rectangle of height 3. (“Exact volume” means no calculator numbers.)

Question

Consider the functions  f(x)=x^3/(x^4+1) and g(x)=x^2/(x^4+1). Let R denote the region in the first quadrant bounded by the curves y=f(x) and y=g(x).
Find the exact volume of the solid that has R as its base if every cross section by a plane perpendicular to the x-axis is a rectangle of height 3. (“Exact volume” means no calculator numbers.)

anonymous · Answer

Okay, so to start off you have to find the area of each cross section, which are rectangles with a height of 3. What would the width of those rectangles be?

amistre64 · Answer

f(x)-g(x)

amistre64 · Answer

the volume of a prism (2 like bases seperated by a given distance)  is just:

 base area * height

anonymous · Answer

Would that be $$3\int\limits_{0}^{1}(x^2/(x^4+1))-(x^3/(x^4+1))dx$$

amistre64 · Answer

assuming your got the intersection point valid then

yep, and if its a negative value you get then just ignore the sign

anonymous · Answer

So now I have to find the antiderivatives of f(x) and g(x). Here is where I keep getting stuck. I can easily find the antiderivative of g(x) to be ln(x^4+1). But I cannot seem to find a way to integrate f(x). Should I not try to find the antiderivatives separately?

amistre64 · Answer

1/4 ln(x^4+1)

anonymous · Answer

How's that?

amistre64 · Answer

take the derivative and see ....

anonymous · Answer

Oh wait that for g(x). That was a typo on my part.

amistre64 · Answer

f(x) is just ugly no matter what if the wolf aint lying to me

amistre64 · Answer

i cant see a simple way to approach it, of course im prolly out of practice on my trig substitutions

anonymous · Answer

So from what I'm getting there's no way to solve this without something they have not taught me in my online class. I'm starting to think that it's a typo (it wouldn't be the first time) and that's supposed to be an x.

amistre64 · Answer

http://www.wolframalpha.com/input/?i=integrate+%28x%5E2-x%5E3%29%2F%28x%5E4%2B1%29 good luck with that ....

amistre64 · Answer

x^2 is spose to be an x instead?

amistre64 · Answer

d/dx tan^(-1)(x^2) = 2x/(x^4+1)  so we are only off by a factor of 2 if thats the case

anonymous · Answer

I think I can take it from here. Thanks for the help

amistre64 · Answer

yep :)