# Thread:

1. ## HELP!! 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.

Edit: Any help at all will be very appreciated.

2. 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

3. ## Solid with parabolic cross-section

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
Attached Thumbnails

4. 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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