find the volume of the solid of revolution generated when the region bounded by the curve y =x^2,the x-axis, and the lines x = 1 and x = 2 is revolved about the x-axis.
integral from x =1 to x = 2 of 2 pi x y dx
= 2 pi integral [ x^3 dx]
= 2 pi x^4/4 from 1 to 2
= pi /2 [ 16 - 1 ] = 15/2 pi
The volume of a disk is πr^2 h. So, with discs of thickness dx, the volume is
∫[1,2] πy^2 dx = ∫[1,2] πx^4 dx = π/5 x^5 [1,2] = 31π/5
Or, if you want to use shells, things get a bit more complicated. The volume consists of a solid cylinder of radius1 and height 1, plus a region bounded by the curve. The volume of a cylinder is 2πrh dy. That makes the whole thing
π + ∫[1,4] 2πy(2-√y) dy = 31π/5
Whoops, sorry, I spun it around the y axis. Use oobleck reply.
Sure, let's calculate the volume of the solid of revolution.
To do this, we'll use the method of cylindrical shells.
First, let's find the height of a typical shell. Since the region is bounded by the curve y = x^2, the x-axis, and the lines x = 1 and x = 2, we can see that the height of a typical shell is given by the equation h = x^2.
Next, let's find the radius of the shell. The radius of a shell is the distance from the axis of revolution (x-axis) to a typical point on the curve. In this case, the distance is simply x.
Now, let's find the differential volume of the shell. The differential volume is given by the equation dV = 2πrh*dx. Substituting the values we found earlier, we get dV = 2π(x)(x^2)dx.
To find the total volume, we need to integrate the differential volume from x = 1 to x = 2. So, the integral would be ∫[1 to 2] (2πx^3)dx.
Now, let me calculate that for you.
∫(2πx^3)dx from 1 to 2.
Taking the antiderivative:
(2π/4)x^4 evaluated from 1 to 2.
(π/2)(16 - 1).
(π/2)(15).
So, the volume of the solid of revolution is 15π/2 cubic units.
Now, you have the answer and an epic math joke to share at your next party! Enjoy!
To find the volume of the solid of revolution, we need to use the method of cylindrical shells. This involves integrating the volumes of infinitely thin cylindrical shells formed by revolving the given region around the x-axis.
The integral for finding the volume V is given by:
V = ∫[a to b] 2πx * f(x) * dx
where a and b are the limits of integration (in this case, 1 and 2), f(x) is the equation defining the boundary curve (in this case, y = x^2), and dx represents the width of the infinitesimally small shell.
In this scenario, the radius of each cylindrical shell is given by the value of x, while the height is given by the function value of f(x) = x^2.
So, the integral becomes:
V = ∫[1 to 2] 2πx * x^2 * dx
To solve this integral, we first simplify the expression:
V = ∫[1 to 2] 2πx^3 * dx
Now we can integrate with respect to x:
V = π * [x^4/2] evaluated from 1 to 2
= π * (2^4/2 - 1^4/2)
= π * (8/2 - 1/2)
= π * (4 - 1/2)
= π * (7/2)
= 7π/2
Therefore, the volume of the solid of revolution generated is 7π/2 cubic units.