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Find the volume of the resulting solid if the region under the curve y = 6/(x² - 5x + 6) from x = 0 to x = 1 is rotated about the x-axis and the y-axis.

User Hkm Sadek
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Final answer:

To find the volume of the resulting solid, you need to integrate the formula for the cross-sectional area about the x-axis and the y-axis. The formula for the cross-sectional area is A = pi * y^2. For rotation about the x-axis, integrate the area formula from x = 0 to x = 1. For rotation about the y-axis, integrate the area formula from y = 0 to y = infinity.

Step-by-step explanation:

To find the volume of the resulting solid, we need to integrate the formula for the cross-sectional area about the x-axis and the y-axis. The formula for the cross-sectional area is A = pi * y^2, where y is the function y = 6/(x^2 - 5x + 6). For rotation about the x-axis, we integrate the area formula from x = 0 to x = 1. For rotation about the y-axis, we integrate the area formula from y = 0 to y = infinity.

For rotation about the x-axis:

Volume = ∫(pi * y^2) dx = ∫(pi * (6/(x^2 - 5x + 6))^2) dx

For rotation about the y-axis:

Volume = ∫(pi * x^2) dy = ∫(pi * x^2 * (dy/dx)) dx = ∫(pi * x^2 * ((dx/dy)^-1)) dx = ∫(pi * x^2 * (1/(dy/dx))) dx = ∫(pi * x^2 * (1/((dy/dx)/(dx/dy)))) dx = ∫(pi * x^2 * (1/(((d/dy)(6/(x^2 - 5x + 6)))/(d/dx)(6/(x^2 - 5x + 6)))) dx = ∫(pi * x^2 * (1/((6 * (2x - 5))/((x^2 - 5x + 6)^2)))) dx = ∫(pi * ((x^2 - 5x + 6)^2)/(6 * (2x - 5))) dx

User Rob Bazinet
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