Find the volume of the solid when the region enclosed by the curves y=((25-(x^2))^(1/2))and y=3 is revolved about the x-axis.
I already have the answer = (256*pi)/3, but don't know how to set up.
From x=-4 to x=4, the y=[(25-(x^2))^(1/2)] curve is above y > 3.
From x = -5 to -4 and 4 to 5, y <3
It is not clear what the x limits of integration are, nor how the two different regions, outside and inside y=3, are to be treated.
I set both y equal to each other to find the points of integration.
squared both sides
(((25-(x^2))^(1/2))=3)^2
then becomes
(x^2)=25-9
Result
x= +(14^(1/2) and -(14^(1/2)
I don't know what to use, as in the disk or washer method.
First find the volume generated when the circle
x^2 + y^2 = 25 from -4 to 4 is rotated (from symmetry I will go from 0 to 4 and double)
V = 2π[integral] (25 - x^2)dx from 0 to 4
= 2π [25x - (1/3)x^3] from 0 to 4
= 2π(100 - 64/3) = 472π/3
then we will "hollow out" a cylinder with radius 3 and height of 8 (from -4 to 4)
that volume is π(9)(8) = 72π
The volume of our rotated solid is
472π/3 - 72π
= 472π/3 - 216π/3
= 256π/3 as required
Thanks
Helped a lot
Congrats to Reiny for that; I was hoping he would take it. I hope you see my point that there are two additional volumes between the two curves from x = -5 to -4 and from 4 to 5. Those volumes are cylindrical on the OUTside. Apparently they expected you to ignore them
To find the volume of the solid formed by revolving the region enclosed by the curves y = (25 - x^2)^(1/2) and y = 3 about the x-axis, you can use the method of cylindrical shells.
First, you need to determine the bounds of integration. To find the points of intersection between the two curves, set them equal to each other:
(25 - x^2)^(1/2) = 3
Squaring both sides, you get:
25 - x^2 = 9
Rearranging, you have:
x^2 = 16
Taking the square root of both sides, you get:
x = ±4
The region enclosed by the curves is bounded by the x-values -4 and 4.
Next, consider a thin horizontal strip of width Δx at some x-value within the bounds of integration. When this strip is revolved about the x-axis, it forms a cylindrical shell with radius y and height Δx.
The radius of the cylindrical shell is the y-coordinate of the curve (25 - x^2)^(1/2), which is the same as the function itself. So, the radius is given by:
r = (25 - x^2)^(1/2)
The height of the cylindrical shell is Δx.
The volume of the cylindrical shell is given by:
V = 2πrhΔx
Substituting the expressions for r and h, the volume can be written as:
V = 2π(25 - x^2)^(1/2)Δx
To find the total volume of the solid, you need to sum up all the volumes of the cylindrical shells from x = -4 to x = 4. This can be represented as a definite integral:
V = ∫[from -4 to 4] 2π(25 - x^2)^(1/2) dx
Evaluating this integral will give you the volume of the solid. You can then simplify the expression and calculate the numerical value, which is (256π)/3, as you mentioned in your question.