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anonymous
 5 years ago
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 xaxis?
anonymous
 5 years ago
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 xaxis?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0first we should determine our bounds right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first draw a rough of a parabola (x^26x+10) and dissect it with a diagonal line (y=x). then set x=x^26x+10. Solve this it would tell you where parabola and the line intersect.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yep.... or x^2 = 7x +10 (x5)(x2)......[2,5]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well it looked right in me head lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00 = x^2 7x +10...better

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0pi [S] [x^2 7x +10]^2 dx [2,5] right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[F(x) = \pi \int\limits_{2}^{5}[x^2 7x +10]dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0F(5)  F(2) = answer of the volume of the solid formed....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0forgot the ^2 in that equation..... there should be an edit button lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0or should we integrate each 'y=' seperately on the interval and then just subtract the higher from the lower?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think you integrate the given parabola? Must check on this.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes you integrate the given function. The other created function is helpful only to create the boundaries.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we 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 :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0piR^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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0but I would feel safer just getting one solid and subtracting the other from it ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure, If you are a bright student you can do several methods to check, but a nonmath 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.
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