Volume Integration: Find Solutions

[blumfeld0]

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

 

[HallsofIvy]

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 $\pi$ in $\pi r^2$! Your integral should be
$$\pi \int_3^4 [(\frac{1}{x^3}+3)^2- 9]dx$$

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 $\pi$ again!

is that even close to right?

thank you

 

[tacman]

For the first problem, your set up for the integral is correct. However, there are a few errors in the expression inside the integral. It should be (-1/x^3)^2 instead of (1/x^3)^2 and the limits of integration should be -3 to -4 since the region is being rotated around the line x=-3. The final integral should be:

V = ∫ from -3 to -4 of [(3^2) - (-1/x^3)^2] dx

For the second problem, your set up is also correct. However, the expression inside the integral should be (x^2)^2 instead of (3-x^2)^2 and (y^2)^2 instead of (3-Sqrt(x))^2. Also, the limits of integration should be from 0 to 1 since the region is being rotated around the line x=-3. The final integral should be:

V = ∫ from 0 to 1 of [(x^2)^2 - (y^2)^2] dy

Overall, your approach to setting up the integrals is correct, but make sure to double check the expressions and limits of integration to ensure the correct solution.