The discussion focuses on calculating the volume of solids of revolution using the washer method and the shell method. The first problem involves finding the volume of the solid formed by rotating the area between the curves y = x² + 1 and y = 9 - x² about the line y = -1, with the correct integral setup being V = π∫[(10 - x²)² - (x² + 2)²] dx from -2 to 2. The second problem requires setting up the integral for the volume of the solid formed by rotating the area between y = tan(x) and y = x about the y-axis, with the integral expressed as V = 2π∫x(tan(x) - x) dx from 0 to π/3. The importance of accurately identifying inner and outer radii in these calculations is emphasized.
PREREQUISITESStudents studying calculus, particularly those focusing on integral applications in volume calculations, as well as educators seeking to clarify methods for teaching solids of revolution.
Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.Jbreezy said:Homework Statement
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
Homework Equations
y = x^2 +1 , y = 9-x^2: about y = -1
The Attempt at a Solution
I used disks.
Jbreezy said:I said that r = 1 + ((9-x^2)-(x^2 +1 )) = 9 - 2x^2
So my integral all together is
V = ∏ ∫ (9 - 2x^2)^2 dx from [-2,2]
Does this set up look correct.?
Looks OK to me. It would have been helpful to indicate that your using shells.Jbreezy said:Also I have another one I would like to confirm.
This one your supposed to set up the integral not evaluate for the volume of a solid.
y = tanx, y = x, x = pi/3 : about the y axis.
I have
V = 2∏∫ x(tanx - x) dx between 0 and ∏/3
I think it is good but just to be safe.
Mark44 said:Did you draw a sketch? Actually, two sketches would be useful - one of the region defined by the two curves, and another that shows the solid of revolution.
Disks won't work, but washers will. A washer is a disk with a hole in its middle. For a typical volume element, the outer radius goes from the line y = -1 to the curve y = 9 - x2. The inner radius goes from the line y = -1 to the curve y = x2 + 1.
Your radii are correct but the integral isn't. The volume of a washer is ##\pi(R^2 - r^2)(\text{thickness})##.Jbreezy said:Yeah I always draw it. Sorry for the confusion with the other problem not indicating I was using shells.
So what about this. r = inner radius, R = outer radius
r = 1 + ( x^2 + 1) R = 1+ (9-x^2)
So V = ∏∫( x^2 +2)^2 + (10-x^2)^2 dx I think that is it but I have a little trouble with the radius sometimes.