The region in the first quadrant bounded by y=6x^2 , 2x+y=8, and the y-axis is rotated about the line x=-1.
The volume of the resulting solid is:
Please help me set up the integral for this one, I'm not sure how to do it. I've tried ∫[-4/3,1] pi (8-2x-1)^2-(6x^2-1)^2 dx, which I know isn't right but I don't know how to fix it.
Nevermind, I got it:
∫[0,1]2pi(x+1)(8-2x-6x^2)dx = 41pi/3
The graphs intersect at (-4/3,32/3) and (1,6)
We are only interested in x>0, so the volume is
using discs
v = ∫[0,1] π(R^2-r^2) dx
where R=8-2x and r=6x^2
v = ∫[0,1] π((8-2x+1)^2-(6x^2+1)^2) dx = 782π/15
Since we are rotating around the line x = -1, the radius is increased by 1, not reduced.
using shells, we have to divide the region into two parts because of the boundary change at y=6
v = ∫[0,6] 2πrh dy + ∫[6,8] 2πrh dy
where r=y+1 and h=√(y/6) or (8-y)/2
v = ∫[0,6] 2π(y+1)√(y/6) dy + ∫[6,8] 2π(y+1)*(8-y)/2 dy
= 184π/5 + 46π/3 = 782π/15
Never mind. I rotated around the line y = -1.
Well, setting up the integral can be a bit tricky, but don't worry, I'm here to help you with a little bit of humor!
First, let's break it down step by step. We need to find the region in the first quadrant bounded by y = 6x^2, 2x + y = 8, and the y-axis.
Now, let's look at the solid that is formed when this region is rotated about the line x = -1. You can think of it as a little carnival ride where the line x = -1 is the axis, and our function is taking a spin.
To set up the integral, we can use the method of cylindrical shells. We'll integrate the volume of each individual shell as it goes round and round, adding up all the little volumes to get the total volume.
Since we are rotating around x = -1, let's consider a tiny vertical slice at position x. The height of our slice would be (6x^2 - 0), the distance from the function to the y-axis, and the length of our slice would be 2x + 8, the distance from the function to the line 2x + y = 8.
To calculate the volume of each little cylindrical shell, we multiply the circumference (2π) by the height (6x^2) and the thickness DX (a small change in x).
So, the volume of each shell is given by V = 2π(6x^2)(2x + 8) DX.
To find the total volume, we integrate this expression over the interval where our function exists.
∫[a,b] 2π(6x^2)(2x + 8) DX
Now, to find the limits of integration, we need to determine where our region begins and ends. Looking at the equation 2x + y = 8, we can solve for x to find x = (8 - y)/2. Since it's bounded by the y-axis, our lower limit is x = 0.
The upper limit is the x-value where our function intersects 2x + y = 8. So we set 6x^2 = 8 - 2x and solve for x. This gives us x = 2/3.
Finally, we have our integral:
∫[0,2/3] 2π(6x^2)(2x + 8) DX
Now, you can simplify and evaluate this integral to find the volume of the resulting solid. Happy calculating, and enjoy the carnival ride of math!
To set up the integral for finding the volume of the solid, we will use the method of cylindrical shells.
First, let's sketch the given region in the first quadrant.
The region is bounded by three curves: y = 6x^2, 2x + y = 8, and the y-axis.
The curve y = 6x^2 is a parabola that opens upward and passes through the origin (0, 0).
The line 2x + y = 8 can be rewritten as y = -2x + 8. When y = 0, we have x = 4. So, the line intersects the x-axis at (4, 0).
To find the limits of integration, we need to determine the bounds along the x-axis.
1. First, find the x-coordinate of the point where the parabola intersects the line.
6x^2 = -2x + 8
6x^2 + 2x - 8 = 0
3x^2 + x - 4 = 0
Factoring or using the quadratic formula, we find x = -4/3 or x = 1.
Since we are rotating the region about the line x = -1, we need to shift the region 1 unit to the right. So, the new bounds become x = -4/3 + 1 = -1/3 and x = 1 + 1 = 2.
2. Next, we need to find the height at each x-coordinate to determine the range along the y-axis.
The top curve is given by y = -2x + 8, and the bottom curve is y = 6x^2.
Now, let's set up the volume integral using cylindrical shells:
V = ∫[a,b] 2πrh dx
Where,
a = the lower bound of x
b = the upper bound of x
r = distance from the axis of rotation (x = -1) to the shell (in this case, r = x + 1)
h = height of the shell (in this case, h = top curve - bottom curve)
Therefore, the volume integral becomes:
V = ∫[-1/3, 2] 2π(x + 1)(-2x + 8 - 6x^2) dx
Evaluating this integral will give us the volume of the resulting solid.