Integral of volume help! 1. Find the volume of the solid obtained by rotating th

Question

Integral of volume help! 1. Find the volume of the solid obtained by rotating the region bounded by the curves about the y-axis... (15y^2)/pi + 25y/pi - 10/pi = [5x/(pi^2)] [arccos(-1)]

Explanation / Answer

Note that y = x^6 and y = 1 (with x > 0) intersect when x = 1. Keeping the functions in terms of x and revolving the region about y = -3, which is parallel to the x-axis, means that we can find the volume by the disc and washer method. Since 1 >= x^6 for x in [0,1], the bigger radius will be 1 - (-3) = 4, while the smaller radius will be x^6 - (-3) = x^6 + 3. Hence, the volume is given by integral(x = 0 to 1) [pi * 4^2 - pi * (x^6 + 3)^2] dx = pi * integral(x = 0 to 1) [16 - (x^12 + 6x^6 + 9) dx. = pi * integral(x = 0 to 1) (7 - x^12 - 6x^6) dx = pi * (7x - x^13/13 - 6x^7/7) for x = 0 to 1. = pi * (7 - 1/13 - 6/7 - 0) = 552 pi/91.

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