Volume integration

  • Thread starter blumfeld0
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hi guys. i just need help setting up the integral for the following two problems.

1. find the volume of the solid obtained by rotating the region by the given area about the specified lines.
y= 1/ x^3, y= 0, x=3, x=4 ABOUT x=-3

so i have integral from 3 to 4 of [(-3)^2 - (1/x^3)^2] dx

is that right?


2. find the volume of the solid obtained by rotating the region bounded by the given curves about specified line
y=x^2, x=y^2 about the line x=-3

so i have integral from 0 to 1 of [(3-x^2)^2 - (3-Sqrt(x))^2] dx


is that even close to right?

thank you
 

Answers and Replies

  • #2
HallsofIvy
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blumfeld0 said:
hi guys. i just need help setting up the integral for the following two problems.

1. find the volume of the solid obtained by rotating the region by the given area about the specified lines.
y= 1/ x^3, y= 0, x=3, x=4 ABOUT x=-3

so i have integral from 3 to 4 of [(-3)^2 - (1/x^3)^2] dx

is that right?
No, but it's close. For each x, you are looking at the region between two circles. The inner circle has radius 3 and the outer circle has radius y+3. Also you have forgotten the [itex]\pi[/itex] in [itex]\pi r^2[/itex]! Your integral should be
[tex]\pi \int_3^4 [(\frac{1}{x^3}+3)^2- 9]dx[/tex]



2. find the volume of the solid obtained by rotating the region bounded by the given curves about specified line
y=x^2, x=y^2 about the line x=-3

so i have integral from 0 to 1 of [(3-x^2)^2 - (3-Sqrt(x))^2] dx
Although, because of the symmetry, that will give you the correct answer, since you are rotating around the vertical line x= -3, your "washers" should be going vertically, so every "x" should be "y". Oh, and you forgot [itex]\pi[/itex] again!


is that even close to right?

thank you
 

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