Find the volume of the solid generated by revolving the triangular region with vertices (1,1), (b,1), and (1,h) about:
a) the x-axis
b) the y-axis
If a<1 and b<1 we get
Using discs:
pi*Int(R^2 - r^2) dx [b,1]
R = 1
r = y = (a-1)(x-b)/(1-b) + 1
= pi/3 (a-1)(a+2)(b-1)
Using shells:
2pi*Int(rh)dy [a,1]
r = y
h = 1-x = 1 - [(1-b)(y-1)/(a-1) + b]
= pi/3 (a-1)(a+2)(b-1)
If a or b > 1, then change integration limits and (1-a) -> (a-1), but the answer is the same
Thank you, Wolframalpha!
oh - that's just around the x-axis. Good luck on the y-axis. By symmetry, the results should look quite similar.
To find the volume of the solid generated by revolving the given triangular region about the x-axis and y-axis, we can use the method of integration.
a) Revolving about the x-axis:
In this case, each vertical strip of the solid will have a radius equal to the distance between the x-axis and the point on the triangle's perimeter. The height of each strip will be the corresponding change in x.
Let's denote the base of the triangle as a, which is equal to "b - 1" since one vertex is located at (1, 1). The height of the triangle is denoted as h.
To find the volume, we need to integrate the area of each strip along the x-axis from x = 1 to x = b.
The area of each strip is given by the formula: A = πr^2, where r is the radius.
Since the radius is equal to the y-coordinate of the triangle's perimeter at each x-value, we have:
r = h - x, where x ranges from 1 to b.
Therefore, the integral for the volume is given by:
V = ∫[1 to b] π(h - x)^2 dx
Expanding and solving the integral will give you the volume V.
b) Revolving about the y-axis:
In this case, each vertical strip of the solid will have a radius equal to the x-coordinate of the point on the triangle's perimeter. The height of each strip will be the corresponding change in y.
Using the same variables as before, the radius is given by:
r = x - 1, where x ranges from 1 to b.
The integral for the volume is then:
V = ∫[1 to b] π(x - 1)^2 dy
Expanding and solving the integral will give you the volume V.
Please let me know if you need the detailed calculations for the integration or any further assistance.
To find the volume of the solid generated by revolving the triangular region about a given axis, we can use the method of disk integration.
a) When revolving the region about the x-axis:
The base of each disk is a circular region that is formed by revolving a line segment (or a slice) of the triangle. The height of each disk is the distance between the x-coordinate of the slice and the x-axis.
- Let's determine the limits of integration for x: The region is bounded by the lines x=1 and x=b. So, the limits of integration for x are 1 to b.
- Since the radius of each disk is given by the function y=f(x) (which represents the distance between the x-axis and the line segment), we need to find the equation of the line that passes through (1, h) and (b, 1). The equation of this line is y = (1 - h)/(b - 1)(x - 1) + h. Therefore, the radius of each disk is the function r(x) = [(1 - h)/(b - 1)(x - 1) + h].
- Using the formula for the volume of a solid of revolution using disk integration, V = π ∫[a,b] (r(x))^2 dx, we can find the volume by integrating the square of the radius over the limits of integration.
b) When revolving the region about the y-axis:
The base of each disk is a circular region that is formed by revolving a vertical line segment (or a slice) of the triangle. The height of each disk is the distance between the y-coordinate of the slice and the y-axis.
- To find the limits of integration, we need to determine the range of y-values for the triangle. From the given vertices, we observe that the vertical range is from y=1 to y=h.
- The radius of each disk is given by the function x=f(y) (which represents the distance between the y-axis and the vertical line segment). Solving the equation of the line passing through (1, 1) and (1, h) for x, we get x = [(b-1)/(h-1)](y-1) + 1. Hence, the radius of each disk is r(y) = [(b-1)/(h-1)](y-1) + 1.
- Using the same formula for the volume of a solid of revolution with disk integration, V = π ∫[c,d] (r(y))^2 dy, we can now calculate the volume by integrating the square of the radius over the limits of integration.
By applying these calculations, you can find the volumes of the solid generated by revolving the triangular region about the x-axis and y-axis, respectively.