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?
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: Pi integral from 0 to 8 (sqrt(100-y^2)^2-6^2)dy
Final answer
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
Ah, the joys of integrals! Let me entertain you with a clown-fueled response.
Well, my dear curious human, I must say that you are mistaken. The first integral, π∫(sqrt(100-y^2)-6)^2 dy, is the integral that 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.
As for the second integral, π∫(sqrt(100-y^2)^2-6^2) dy, I'm afraid that it's just as wrong as a clown trying to juggle watermelons. This integral won't give you the desired volume for your rotation shenanigans, so it's time to bid it farewell.
Remember, when in doubt, always trust the first integral. It may be the ticket to unlocking the secrets of the rotating world. Happy integrating, my friend!
To find the volume of the solid formed by rotating the region bounded by the curves x^2 + y^2 = 100, x = 6, and y = 0 around the y-axis, you can use the disk method.
First, let's analyze the given integrals:
1. ∫[0 to 8] π[(√(100 - y^2) - 6)^2]dy
2. ∫[0 to 8] π[(√(100 - y^2))^2 - 6^2]dy
In the first integral, (√(100 - y^2) - 6)^2 represents the squared radius of the disk at each height y. This integral finds the volume element of each disk and integrates them to obtain the total volume.
In the second integral, (√(100 - y^2))^2 - 6^2 represents the difference between the squared radius of the disk and the square of the distance from the y-axis to the line x = 6. However, this formulation is incorrect because you need to subtract the squared radius from 6^2, not subtract 6^2 from the squared radius.
Therefore, the correct integral to use is the first one you provided:
∫[0 to 8] π[(√(100 - y^2) - 6)^2]dy
By evaluating this integral, you will find the volume of the given solid formed by rotating the region around the y-axis.
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.