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.
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.
recall that the volume between and ( ) is given by , where is the formula for the cross-sectional area.
in this case, our cross-sections are parabolas of the form . 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 for arbitrary . find the intercepts and all that. now find the area under this curve. we see the area will be given by: .
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
Hi -
I attach a drawing of the problem as I understand it, with three axes: , and .
The values of go from to . At , the parabola (which lies in a plane parallel to the plane) has equation . This parabola will have values of from to , and values of from to .
What you need to do, then, to find the volume of the solid enclosed by all these parabolas, the plane and the plane , is:
- Find the area of the typical parabola shown. (Do this in the usual way with an integral, whose limits are to .)
- Replace by in your formula for .
- Now imagine increasing by an amount . As it does so, the volume 'swept out' is approximately . So the total volume will be . So, replace by your formula in terms of , 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