find the volume of the solid obtained by rotating the region in the first quadrant bounded by x=0, y=1, x=y^4 about the line y=10

Question

I just need help with the answer

Answer 1

I used the formula

V=π∫|(f(x)-d)² -(g(x)-d)²|dx Integrating from a to b

For this case, a=0, b = 1, g(x) = 1, f(x) = x^1/4

V = π∫|(x^1/4-10)²-(1-10)²dx = π∫|x^1/2-20x^1/4+100 -81|dx

= π[(2/3)x^3/2-20(4/5)x^5/4+19x] evaluated from at 1 and 0

= π|(2/3 -16 +19) = 11π/3

Answer 2

By Disk Method: V = π∫₀¹[(10 - x^1/4)² - 9²]dx = π∫₀¹(19 - 20x^1/4 + x^1/2)dx = π(19x - 16 x^5/4 + 2/3x^3/2)₀¹ =

π(19 - 16 + 2/3) = 11/3π = 11.52

Answer 3

If I understand the problem, the area is the y curve y = x.25 from x = 0 to x = 1 rotated around the line y = 10

First calculated the y center of area y-bar which equals 5/12. The distance from that point to y = 10 is

10 - 5/12 = 115/12.

The calculated area is 4/5.

The volume rotated around the line y = 10 is 2*pi*(10 - 5/12)(4/5) = (46/3)*pi