Find the volume of the solid generated by revolving the region bounded by the given curves about the x-axis.
y=x^2+1
x+y=3
Please help me set up the integral!
First you need their intersection, that should be easy
Hope you get (-2,5) and (1,2)
The outer radius is (3-x)
the inner radius is (x^2+1)
so you want ∫ ( (3-x)^2 - (x^2+1)^2 )dx from x = -2 to 1
expand and the rest is pretty straightforward.
Let me know what you get
Awesome, that's what I got!
To find the volume of the solid generated by revolving the region bounded by the curves about the x-axis, we can use the method of cylindrical shells.
First, let's sketch the region bounded by the given curves:
The curve y = x^2 + 1 is a parabola that opens upwards and has its vertex at (0,1). The line x + y = 3 is a straight line with an x-intercept of 3 and a y-intercept of 3.
To find the points of intersection between the curves, we can set them equal to each other:
x^2 + 1 = 3 - x
Rearranging the equation, we get:
x^2 + x - 2 = 0
Factoring the quadratic equation, we have:
(x + 2)(x - 1) = 0
So, x = -2 or x = 1.
Now, let's set up the integral to find the volume of the solid:
Since we are revolving the region bounded by the curves about the x-axis, we integrate with respect to x.
The height of the cylindrical shell at each x-value is given by the difference between the two curves: (3 - x) - (x^2 + 1).
The radius of the cylindrical shell is equal to x, as it is the distance from the axis of rotation (x-axis) to the curve.
The differential element of volume, dV, is therefore given by:
dV = 2πx[(3 - x) - (x^2 + 1)] dx
To find the total volume, we integrate dV from the lower bound x = -2 to the upper bound x = 1:
V = ∫[from -2 to 1] 2πx[(3 - x) - (x^2 + 1)] dx
Now, you can evaluate this integral to find the volume of the solid generated by revolving the region bounded by the given curves about the x-axis.