Find the volume if the region enclosing y=1+ sqrerootof x, x=0, y=0, and x=9 is rotated about the x axis.
I used the disk method and came up with the integral: pi integral (from 0-9) of (1+sqrerootx)^2 dx, but my answer was wrong.
how did you integrate (1+√x)^2 ?
I would suggest you expand it first to get
1 + 2√x + x
when you integrate that you get
x + (4/3)x^(3/2) + (1/2)x^2
so my arithmetic shows
vol = pi(9 + (4/3)(27) + 81/2)
= .....
Oh...thanks Reiny, I threw it into my calculator wrong (forgot the parenthesis at the beginning of the function)
To find the volume of the region enclosed by the curves y = 1 + √x, x = 0, y = 0, and x = 9, when rotated about the x-axis, you are correct in using the disk method.
However, the integral you wrote seems to have an error with the term (1 + √x)^2. Let's correct it together.
First, let's find the limits of integration. We know the region is bounded by x = 0 and x = 9.
To find the volume using the disk method, we need to consider the cross-sections formed by rotating the region. Each cross-section is a disk with a radius determined by the perpendicular distance from the x-axis to the curve at a given x-value.
At a given x, the radius of a cross-section is given by y = 1 + √x. However, since we are rotating about the x-axis, the perpendicular distance is the y-value itself, not the absolute value.
Therefore, the radius of each disk is simply √x.
To calculate the volume, we need to integrate the areas of these disks over the interval [0, 9].
The integral for the volume is given by:
V = π ∫[0,9] (r(x))^2 dx
= π ∫[0,9] (√x)^2 dx
= π ∫[0,9] x dx
Evaluating this definite integral, we get:
V = π [x^2/2] from 0 to 9
= π [9^2/2 - 0^2/2]
= π [81/2]
= 40.5π
Therefore, the volume of the region enclosed by the curves y = 1 + √x, x = 0, y = 0, and x = 9 when rotated about the x-axis is 40.5π cubic units.