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Let R be the triangular region in the first quadrant bounded by y = x, y = -x + 8, y = 2

a) Compute the volume of the solid obtained by revolving R around the y-axis in two ways: the method of cylindrical shells and the washer method
b) Compute the volume of the solid obtained by revolving R around the x-axis in two ways: the method of cylindrical shells and the washer method

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Final answer:

To compute the volume of the solid obtained by revolving the region R around the y-axis and the x-axis, we can use the methods of cylindrical shells and washers. For the y-axis, the integral becomes ∫[0,2] 2πx(8-2x) dx for shells and ∫[0,2] (π[(8-x)² - x²)] dx for washers. For the x-axis, the integral becomes ∫[0,8] 2πy(8-2y) dy for shells and ∫[0,8] (π[(8-y)² - y²)] dy for washers.

Step-by-step explanation:

To compute the volume of the solid obtained by revolving the region R around the y-axis, we can use the method of cylindrical shells. The volume is given by the integral of the product of the circumference of the shell, the height of the shell, and the thickness of the shell. We integrate from y = 0 to y = 2, and the shell height is given by the difference between the top and bottom curves, which are y = -x + 8 and y = x. So, the integral becomes ∫[0,2] 2πx(8-2x) dx. Similarly, for the washer method, we integrate the area of each washer, which is given by the difference between the outer and inner radii squared times π. The outer radius is y = 8 - x and the inner radius is y = x. So, the integral becomes ∫[0,2] (π[(8-x)² - x²)] dx.

To compute the volume of the solid obtained by revolving the region R around the x-axis, we can follow the same steps as before, but with the roles of x and y reversed. For the method of cylindrical shells, we integrate from x = 0 to x = 8, and the shell height is given by the difference between the top and bottom curves, which are x = 8 - y and x = y. So, the integral becomes ∫[0,8] 2πy(8-2y) dy. For the washer method, we integrate the area of each washer, which is given by the difference between the outer and inner radii squared times π. The outer radius is x = 8 - y and the inner radius is x = y. So, the integral becomes ∫[0,8] (π[(8-y)² - y²)] dy.

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