Question

1) A region is bounded by the line y = x and the parabola y = x2 - 6x + 10. What is the volume of the solid generated by revolving the region about the x-axis?

Answer 1

first we should determine our bounds right?

Answer 2

first draw a rough of a parabola (x^2-6x+10) and dissect it with a diagonal line (y=x). then set x=x^2-6x+10. Solve this it would tell you where parabola and the line intersect.

Answer 3

yep....

or x^2 = -7x +10

(x-5)(x-2)......[2,5]

Answer 4

well it looked right in me head lol

Answer 5

0 = x^2 -7x +10...better

Answer 6

pi [S] [x^2 -7x +10]^2 dx [2,5] right?

Answer 7

\[F(x) = \pi \int\limits_{2}^{5}[x^2 -7x +10]dx\]

Answer 8

F(5) - F(2) = answer of the volume of the solid formed....

Answer 9

forgot the ^2 in that equation..... there should be an edit button lol

Answer 10

or should we integrate each 'y=' seperately on the interval and then just subtract the higher from the lower?

Answer 11

I think you integrate the given parabola? Must check on this.

Answer 12

Yes you integrate the given function. The other created function is helpful only to create the boundaries.

Answer 13

we tend to get an area shaped like this right?

if we find the volume of the solid formed by the y=x in the interval and then subtract the volume of the solid formed by y=x^2 -6x +10...we will end up with the volume of the solid of this area..... and from the last time i messed this up, I recall we had to use pi(y^2) as the intgrands :)

Answer 14

piR^2 - pir^2 = pi(R^2 - r^2) so we should be able to use the (y=x)^2 - (y=x^2....)^2 as the integrands

Answer 15

but I would feel safer just getting one solid and subtracting the other from it ;)

Answer 16

Sure, If you are a bright student you can do several methods to check, but a non-math major: to find the volume of a solid integrate from a to b, pi*(f(x))^2. (f(x) is given function. Find a, b boundaries as we did above.