the base of a solid is the region bounded by the parabolas y=x^2 and y=2-x^2..?

Question

the base of a solid is the region bounded by the parabolas y=x^2 and y=2-x^2. find the volume of the solid if the cross-sections perpendicular to the x-axis are squares with one side lying along the base.

would someone please help me with this problem?...its confusing >.<
thanx a lot!! ^ ^

Answer 1

This one is hard to visualize, but once you get the picture, there's a lot of symmetry, but you have to set it up very carefully. After that, the integration is easy.

Start by drawing those two parabolas in the xy plane. Notice that the region is symmetric about the y-axis, and also about the line y=1.

That means the base can be divided into four symmetric regiions, and one of them is bounded by y = x^2 and by the lines x=0 and y=1.

Now about " the cross-sections perpendicular to the x-axis." If cross-sections are perpendicular to the x-axis, they're in or parallel to the yz plane. And these cross-sections are squares.

In your diagram of the two parabolas, draw a narrow differential element vertically in the first quadrant from y=x^2 up to the line y=1. Draw it just like you're going to integrate on dx -- which you will do in a minute.

This element has length 1 - x^2, and it represents the lower half of the base of a square. Double that, and you get the y-distance between the two parabolas.

So the base of your square is 2(1-x^2). If that's the base, then the entire square is 4(1-x^2)^2.

Now you're going to integrate on x. But because of symmetry about the y-axis, you can integrate on x from 0 to 1, and double the result to get your answer.

So your integral is 2 [4 (1-x^2)^2] dx. You can expand that expression to get 8(1 - 2x^2 + x^4) dx, and when you integrate, you get 8 [x - (2/3)x^3 + (1/5)x^5] evaluated from 0 to 1.

The lower limit is zero, and the upper limit is

8(1 - 2/3 + 1/5) = 64/15 (ANSWER)

Answer 2

pi * (integral [top function - bottom function]

evaluate at the intersection points

I'd explain better but I just wrote a Calculus exam today, and I never have to do it agaiN! not this year at least!

Answer 3

HH