Find the volume of the solid formed when the region bounded by y=3x^2 and y=36-x^2 is revolved about the x-axis.
I see several posts made by you dealing mainly with these integrations problems, without any indication by you showing any steps or attempts.
I will do this one. Try to follow these steps to solve the others
Make a sketchl
First you need the intersection to see the region that you are revolving.
y = 3x^2 and y = 36-x^2
3x^2 = 36-x^2
4x^2 = 36
x = ± 3
Because of the symmetry, I will boundaries from 0 to 3, then double our volume
in general,
Volume = π∫y^2 dx from left boundary to right boundary
the y , the height or radius of our rotation, is
36-x^2 - 3x^2 = 36-4x^2
V = 2π∫ (36-4x^2)^2 dx from 0 to 3
= 2π∫ (1296 - 288x^2 + 16x^4) dx from 0 to 3
= 2π [ 1296x - 96x^3 + (16/5)x^5] from 0 to 3
= 2π (3888 - 2592 + 3888/5 - 0 - 0 - 0)
= (20736/5)π = appr. 13028.8
I use this to check my answers. I apologize for not realizing that I should have proven to you that I was doing them by myself at first.
Fair enough.
In that case, you should post the answer you got, that way I can just do some fast scribbling on a piece of scrap paper and either confirm or reject your answer.
It would save us a lot of unnecessary typing of the whole solution.
To find the volume of the solid formed by revolving the region between two curves around the x-axis, we can use the method of cylindrical shells.
First, let's find the points of intersection between the two curves y = 3x^2 and y = 36 - x^2. By setting the two equations equal to each other, we have:
3x^2 = 36 - x^2
Combining like terms, we get:
4x^2 = 36
Dividing both sides by 4, we have:
x^2 = 9
Taking the square root of both sides, we get:
x = ±3
So, the two curves intersect at x = -3 and x = 3.
To set up the integral for the volume of the solid, we need to express the height of each infinitesimally small cylindrical shell. The height of each shell is given by the difference between the upper curve and the lower curve at a particular x-value.
The upper curve is y = 36 - x^2 and the lower curve is y = 3x^2.
Therefore, the height of each shell, h, is given by h = (36 - x^2) - (3x^2) = 36 - 4x^2.
Next, we need to determine the radius of each infinitesimally small cylindrical shell. Since we are revolving the region around the x-axis, the radius is simply the x-value.
The radius, r, is given by r = x.
Now, we can set up the integral to find the volume V:
V = ∫[a, b] 2πr * h dx
Where a and b are the x-values of the region of interest. In this case, a = -3 and b = 3.
V = ∫[-3, 3] 2πx * (36 - 4x^2) dx
Evaluating this integral will give us the volume of the solid formed when the region is revolved about the x-axis.