Compute the volume of the solid obtained by rotating the region in the first quadrant encolsed by the graphs of the functions y = x^2 and y = sqrt(x) about the y-axis.
I keep getting (1/3)pi and can't figure out what I'm doing wrong!
rotating about the y -axis needs
V = π∫x^2 dy
first equation:
y = x^2
2nd equation:
√x = y
x = y^2
x^2 = y^4
so V = π∫y dy - π∫y^4 dy
= π [ (1/2)y^2 - (1/5)y^5] from y = 0 to 1
= π( 1/2 - 1/5)
= (3/10)π
To compute the volume of the solid obtained by rotating the region in the first quadrant, enclosed by the graphs of the functions y = x^2 and y = sqrt(x), about the y-axis, you need to use the method of cylindrical shells.
First, let's graph the two functions to get a better understanding of the region that needs to be rotated:
Graph of y = x^2:
This is a parabolic curve opening upwards and passing through the origin (0,0).
Graph of y = sqrt(x):
This is a square root curve that starts from the origin (0,0) and increases as x increases.
Now, to find the volume, we need to integrate the volume of each cylindrical shell that makes up the solid.
The volume of a cylindrical shell can be calculated using the formula: V = 2πrhΔx, where r is the radius of the shell, h is the height of the shell, and Δx is the thickness of the shell.
In this case, we are rotating the region around the y-axis, so the radius of each cylindrical shell is x (the distance from the y-axis), and the height of each shell is the difference between the two functions: (x^2 - sqrt(x)).
To set up the integral, we need to determine the limits of integration. Since the region is in the first quadrant, we can integrate from x = 0 to x = 1, which is the point of intersection between the two functions.
Now, let's set up the integral:
V = ∫[0 to 1] 2πx(x^2 - sqrt(x))dx
To compute this integral:
1. Expand the expression inside the integral:
V = ∫[0 to 1] 2π(x^3 - xsqrt(x))dx
2. Integrate term by term:
V = 2π ∫[0 to 1] (x^3 - xsqrt(x))dx
= 2π [(x^4/4) - (2/3)x^(3/2)] evaluated from 0 to 1
= 2π [(1/4) - (2/3)]
3. Calculate the result:
V = 2π [(1/4) - (2/3)]
= 2π [(3/12) - (8/12)]
= 2π [-5/12]
= - (5/6)π
It seems like you made a sign error when integrating the term with the square root function. Double-check your calculations, and you should get the correct answer of -(5/6)π, which is approximately -0.833π. Remember to consider the appropriate limits of integration to capture the desired region.