Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=2x+1 about the x-axis over the interval [0,1]. I have no clue how to start this..... :(
To find the volume of the solid obtained by rotating the region under the graph of the function f(x) = 2x + 1 about the x-axis, you can use the method of cylindrical shells.
Here's how you can approach this problem step by step:
1. First, let's graph the function f(x) = 2x + 1 over the interval [0, 1]. This will give you a straight line passing through the points (0, 1) and (1, 3) on the coordinate plane.
2. Next, imagine the region bounded by this graph, the x-axis, and the vertical lines x = 0 and x = 1. This region looks like a right-angled triangle.
3. To find the volume of the solid obtained by rotating this region about the x-axis, you need to consider an infinitesimally thin cylindrical shell with radius r and height h.
4. The radius of each shell is the distance of the slice from the axis of rotation, which is x.
5. The height of each shell is the difference in y-values between the function and the x-axis, which is f(x) - 0 = f(x) = 2x + 1.
6. The circumference of each shell is given by 2πr = 2πx.
7. With the equation for the volume of a cylindrical shell as V = 2πr * h, substitute in the values for r and h obtained in steps 4 and 5.
8. Thus, the volume of the solid is V = ∫[0,1] (2πx) * (2x + 1) dx.
9. Evaluate the integral ∫(2πx)(2x + 1) dx over the interval [0, 1] to find the volume of the solid.
10. Finally, integrate the given function over the given interval using methods such as the power rule and properties of integration.
Once you've evaluated the integral, you will have found the volume of the solid obtained by rotating the region under the graph of f(x) = 2x + 1 about the x-axis over the interval [0, 1].
For you to say that you have no clue how to start this straightforward question is not a good sign if you are in a Calculus course.
by Calculus, take vertical discs of width dx and height 2x+1
V = π ∫ r^2 dx
V = π∫(2x+1)^2 dx from 0 to 1
= π [ (1/6)(2x+1)^3 ] from 0 to 1
= π ((1/6)(27) - (1/6)(1^3) )
= π ( 27/6 - 1/6)
= 13π/3