Which of the following integrals correctly computes the volume formed when the region bounded by the curves x^2 + y^2 = 100, x = 6, and y = 0 is rotated around the y-axis?

Question

I've narrowed it down to either

pi∫(sqrt(100-y^2)-6)^2 dy or

pi∫(sqrt(100-y^2)^2-6^2) dy

where a=0 and b=8
Im thinking the second one.

Answer 1

well you know that if you think of the solid as a stack of washers of thickness dy,

v = ∫[0,8] π(R^2-r^2) dy
where R=x=√(100-y^2) and r =6. So,
v = ∫[0,8] π(100-y^2-36) dy

Looks like your 2nd choice above.

Answer 2

lkjgc

Answer 3

The region being rotated is the triangular slice with vertices at ... . Recall that for washers, the volume is
v = ∫ π(R^2-r^2) dy
Here, R is the circle, and r is the line x=6.
Answer: Pi integral from 4 to 0 (sqrt(25-y^2)^2-6^2)dy

Answer 4

Answer: Pi integral from 0 to 8 (sqrt(100-y^2)^2-6^2)dy
Final answer