I didn't know how to explain the problem in words so I made a scan of it.

http://i18.photobucket.com/albums/b1...1/mathprob.jpg

Edit: Any help at all will be very appreciated.

- Dec 9th 2008, 06:24 AMsj9110HELP!! Volume of a solid w/ uniform cross sections of a parabola
I didn't know how to explain the problem in words so I made a scan of it.

http://i18.photobucket.com/albums/b1...1/mathprob.jpg

Edit: Any help at all will be very appreciated. - Dec 9th 2008, 07:38 AMJhevon
recall that the volume between $\displaystyle x = a$ and $\displaystyle x = b$ ($\displaystyle a \le b$) is given by $\displaystyle V = \int_a^b A(x)~dx$, where $\displaystyle A(x)$ is the formula for the cross-sectional area.

in this case, our cross-sections are parabolas of the form $\displaystyle y = a - x^2$. but what is the area of this parabola that fits your constraints? recall that the area "under" the curve is given by the integral. so graph the curve $\displaystyle y = a - x^2$ for arbitrary $\displaystyle a$. find the intercepts and all that. now find the area under this curve. we see the area will be given by: $\displaystyle A = \int_{\sqrt{a}}^{\sqrt{a}} (a - x^2)~dx = 2 \int_0^{\sqrt{a}}(a - x^2)~dx$.

Now, when you get the integral, replace the a with x, since in this case, a = x (the parabola is bounded by y = x, so the height is given by x, so that a = x since a is the height of our parabola above the x-axis). then go to the volume formula i gave you above to find the answer. x ranges from 0 up to 10 - Dec 9th 2008, 07:50 AMGrandadSolid with parabolic cross-section
Hi -

I attach a drawing of the problem as I understand it, with three axes: $\displaystyle Ox$, $\displaystyle Oy$ and $\displaystyle Oz$.

The values of $\displaystyle x$ go from $\displaystyle 0$ to $\displaystyle 10$. At $\displaystyle x = a$, the parabola (which lies in a plane parallel to the $\displaystyle y-z$ plane) has equation $\displaystyle z=a-y^2$. This parabola will have values of $\displaystyle y$ from $\displaystyle -\sqrt a$ to $\displaystyle \sqrt a$, and values of $\displaystyle z$ from $\displaystyle 0$ to $\displaystyle a$.

What you need to do, then, to find the volume of the solid enclosed by all these parabolas, the $\displaystyle x-y $ plane and the plane $\displaystyle x = 10$, is:

- Find the area $\displaystyle A$ of the typical parabola shown. (Do this in the usual way with an integral, whose limits are $\displaystyle y=-\sqrt a$ to $\displaystyle y=\sqrt a$.)
- Replace $\displaystyle a$ by $\displaystyle x$ in your formula for $\displaystyle A$.
- Now imagine $\displaystyle x$ increasing by an amount $\displaystyle \delta x$. As it does so, the volume 'swept out' is approximately $\displaystyle A\delta x$. So the total volume will be $\displaystyle \int_0^{10} A dx$. So, replace $\displaystyle A$ by your formula in terms of $\displaystyle x$, and then work out the integral.

Have I given you enough to go on? Let me know if you want me to check your working.

Grandad - Dec 9th 2008, 07:56 AMJhevon