Isaac H.

asked • 06/06/23

The region ℜ is bounded by ƒ(x) = x^2 - 5x + 9, and y = 5. The region ℜ is rotated about the line x = -1 to obtain the solid S.

The region ℜ is bounded by ƒ(x) = x2 - 5x + 9, and y = 5. The region ℜ is rotated about the line x = -1 to obtain the solid S. We want to find the volume of the solid S using the cylindrical shell method. Please answer the following questions:


a) Sketch the region ℜ in the xy plane.


b) Sketch the cylindrical shell in the xy plane showing and labeling the radius r and the height h of the cylindrical shell.


c) Find the height h of the cylindrical shell.


d) Find the radius r of the cylindrical shell.


e) Find the circumference of the base of the cylindrical shell.


f) Evaluate the volume of the solid S using the above information.

1 Expert Answer

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Doug C. answered • 06/07/23

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Isaac H.

I have a question about parts e) and f) of this problem because I got lost around there. For part e), would the circumference of the base of the cylindrical shell be 2π(x + 1) = 2πx + 2π or something else? For part f), when evaluating the volume of the solid S, will the final answer be (63π/2) after solving the integral?
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06/07/23

Doug C.

Yes to part f, evaluating the integral leads to 64pi/2 (exact answer). For part e) I am not even sure why that question is being asked. The circumference of the base has nothing to do with finding the volume of a particular shell. As a matter of fact it is not clear what is meant by the circumference of the base. The shell has an inner wall and an outer wall. If you call R1 the radius to the outer wall and R2 the radius to the inner wall, then the VOLUME of the shell would be (piR1^2h - piR2^2h). This is the familiar formula for the volume of a right circular cylinder. So volume of outer cylinder minus volume of inner cylinder. Factoring out the pi h leaves pi h (R1^2-R2^2) which can be written as: pi h (R1+R2)(R1-R2). This can be written as: 2pi h (R1+R2)/2 (R1-R2). The (R1+R2)/2 we call the average radius (R) which is the distance of the axis of revolution to the center of the shell. The (R1-R2) is the thickness of the shell (dR). So the volume of a particular shell is: 2pi (height of shell)(average-radius)(thickness) = 2pi f(x) R (dR), which when applied to a Riemann sum becomes the formula used for finding volume by cylindrical shells. Notice that in the above discussion there is no mention of the circumference of the base. Likely the answer for this part of the problem was intended to be 2pi(x+1), but if possible you might want to ask whoever generated this problem, WHY the circumference is being asked for?
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06/07/23

Doug C.

Hopefully when you visited the Desmos graph you started with row and worked your way down to row 21 (the exact answer), perhaps taking notes along the way.
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06/07/23

Isaac H.

Thank you very much for your help and explanation! I can understand this calculus problem now.
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06/08/23

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