Originally posted by joe shmo
Find the volume of a solid, who's base is bounded by the graph of
y=x^3 , y=0, x=1
and whose cross section is that of a semiellipse(a hieght, twice that of it base) taken perpendicular to the y axis.
solving the problem isn't a problem, Im intentionally making it more difficult that it has to be in hopes of gaining a little insigh dy
I dont even relistically think i can?
does this seem logical, or is it horse crap?
a cursory examination of your computation seems like you're being accurate... however, what is your goal in solving the problem this way?
are you trying to take a multivariable problem down to a single parameter? because as you can probably tell it makes the problem a lot harder than it needs to be. this is like finding the volume of a sphere in rectangular coordinates vs. using spherical coordinates, or like stubbornly sticking to cartesian coordinates in a problem better suited to polar.
that being said, i don't think anything you've done is "wrong" except it seems you've arrived at an antiderivative that can't be computed algebraically (as far as i can tell). this doesn't mean it doesn't exist - in fact the integral is definitely possible. you just might have to find the limit of a nasty riemann sum, rather than using algebraic methods of finding an antiderivative and subsequently computing F(1) - F(0).
all in all, i'd rather just use a double integral with respect to x and y. much easier
EDIT: in a look back, i'm not sure why you searched for the quadratic at the beginning of the problem. isn't the problem saying that if you take a cross-section perpendicular to the y-axis (and thus see the relationship between z and x) that the height is twice the length of the base? i.e. wouldn't z = 2x rather than z = -x^2 + 8x? or am i too tired to think straight right now? lol