HELP!! Volume of a solid w/ uniform cross sections of a parabola

**sj9110** · December 9th 2008, 06:24 AM

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

Edit: Any help at all will be very appreciated.

**Jhevon** · December 9th 2008, 07:38 AM

Originally Posted by sj9110

I need to find the volume of a parabola a-x^2 where a is 0 to 10 and its also bounded by y=x or something. I don't know. Clearly I am very confused. I have a drawing of what the solid should look like.

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

Any help at all will be very appreciated.

recall that the volume between $x = a$ and $x = b$ ( $a \le b$ ) is given by $V = \int_a^b A(x)~dx$ , where $A(x)$ is the formula for the cross-sectional area.

in this case, our cross-sections are parabolas of the form $y = a - x^2$ . 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 $y = a - x^2$ for arbitrary $a$ . find the intercepts and all that. now find the area under this curve. we see the area will be given by: $A = \int_{\sqrt{a}}^{\sqrt{a}} (a - x^2)~dx = 2 \int_0^{\sqrt{a}}(a - x^2)~dx$ .

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

**Grandad** · December 9th 2008, 07:50 AM

Hi -

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

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

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

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

**Jhevon** · December 9th 2008, 07:56 AM

Originally Posted by Grandad

Hi -

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

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

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

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

this is what i mean, i answered this problem in the other thread, giving the same solution (though i must admit, Grandad was a bit more elegant, using 3-dimensions and giving a nice diagram to boot--but it was still the same solution).

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HELP!! Volume of a solid w/ uniform cross sections of a parabola

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