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Can explain how to do this calculus question to me?
On a test I had recently it had this question:
Let a>0 and b>0 be given constants. Let R be the region in the first quadrant limited by y=ax^2 and y=b. Find the volume of the solid obtained by rotating R about the y-axis.
I know the answer and I can sketch the region properly but I still don't understand the reasoning as to why my professer used a diffrent function to solve than I did.
3 Answers
- 1 decade agoFavorite Answer
To find the volume of a region rotated about the y-axis, we need the following formula:
V = integral [pi*(radius)^2 dy]
The radius needs to be a function with respect to y. We will call it R(y).
Solving the equation y=ax^2 for x, we get:
x = sqrt(y/a)
We can ignore the negative part of the sqrt, as we are only concerned with values in the first quadrant. This is our radius function, R(y)
V = integral [pi*(R(y))^2 dy]
V = pi integral [(sqrt(y))^2 dy]
V = pi integral [y dy]
The region is bounded by x-axis and y=b, so we must evaluate the integral for from y = 0 to b
V = pi [b^2/2] This is our solution.
Note that had the equation asked for the region to be rotated about the x-axis, we would need to define our equations in terms of x.
It's tempting to want to calculate the area first and try to rotate it. Remember that we are rotating each (different) radius at an increment along y to form a circular area. Plainly speaking, the integration combines these into a volume.
Hope this was helpful!
- 1 decade ago
i would find the area of the curve first using integers then elevate it to ^3. to find the volume of the rotating curve.
