A solid of revolution is formed by rotating the region between the graph of g(y)=
y3

3
and the y-axis over 1

Question

A solid of revolution is formed by rotating the region between the graph of g(y)=
y3

3
  and the y-axis over 1<y<2,   around the y -axis.

Which of the following integrals give the volume of the solid?

BlackTom AI · Accepted Answer

We are given a solid of revolution formed by rotating the region between the graph of g(y) = y^3 and the y-axis over 1 < y < 2, about the y-axis. To determine the volume, visualize a horizontal slice at a fixed y between 1 and 2. The region between x = 0 (the y-axis) and x = y^3 is swept around the y-axis, producing a disk of radius R = x_max = y^3. The cross-sectional area is A(y) = πR^2 = π(y^3)^2 = πy^6. The volume is obtained by integrating these disks with respect to y from y = 1 to y = 2: V = ∫_{1}^{2} π y^6 dy.

Now evaluate or inspect each option:

Option 1: This option shows an integral that appears to involve dx (a dx integral) and includes an expression with (3x) inside the integrand. Since our setup is a rotation about the y-axis using vertical slices in y, the natural variable is y, not x, and the radius involves y^3, not a function of x. Therefore this option mismatches both the correct variable of integration and the correct disk radius expression, so it does not represent the volume.

Option 2: This option includes a different integrand structure, with a √(3x) or similar term and still uses dx. The presence of a square root or an x-variable in the integrand signals a different geometry (likely attempting washers with respect to x) that does not align with the given problem (rotation about the y-axis with horizontal extent from 0 to y^3 for each y). This is not the correct setup.

Option 3: This option shows V = ∫_{2}^{1} π y^3 dy after a rearrangement. First, the limits are reversed (2 to 1 instead of 1 to 2), which would introduce a negative sign or require flipping the limits. Second, the integrand is π y^3, not π y^6, which would correspond to a different radius (y^something) than the actual y^3 radius squared. Thus, this option wrongfully lowers the power and uses improper limits.

Option 4: This option presents V = ∫_{8}^{3} π y^6 dy with odd-looking limits (8 to 3) and an extra factor or spacing that makes the expression unclear. The limits do not match the required 1 to 2 interval, and even though the integrand π y^6 appears correct in form, the limits invalidate the setup. Hence this is not correct.

Option 5: This option shows V = ∫_{2}^{1} π y^6 dy with reversed limits and an extraneous digit placement that seems to reflect a misprint. The structure π y^6 dy is the correct integrand for the disk method about the y-axis with radius y^3, but the limits are swapped (2 to 1 instead of 1 to 2). Because the limits are incorrect, this option does not yield the correct volume.

In summary, the correct conceptual setup requires an integral with respect to y from 1 to 2, integrand πy^6, i.e., V = ∫_{1}^{2} π y^6 dy. The other options either use the wrong variable of integration, employ an incorrect exponent or radius, or have improper limits, all of which produce an incorrect expression for the volume.

A solid of revolution is formed by rotating the region between the graph of g(y)= y3 3 and the y-axis over 1<y<2, around the y -axis. Which of the following integrals give the volume of the solid?

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