Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.
y = 27x3, y = 0, x = 1; about x = 2
Vol = π∫(27x^3)^2 dx from 1 to 2
= π∫ 729x^6 dx from 1 to 2
= π[729/7 x^7] from 1 to 2
= π[ 729/7*128 - 729/7*1]
= ....
your turn
oops, did not read the questions carefully
I simply rotated everything about the x-axis, my mistake.
y = 27x^3
y/27 = x^3
y^(1/3) / 3 = x
The whole solid rotated around the line x = 2 , from y = 0 to y = 27 is
π∫ (2 - x)^2 dy from 0 to 27 , ..... (when x = 1, y = 27)
= π∫ (2 - (1/3)y^(1/3) )^2 dy
= π ∫( 4 - (4/3)y^(1/3) + (1/9)y^(2/3) ) dy
= π[ 4y - y^(4/3) + (1/15)y^(5/3) ] from 0 to 27
= π ( 108 - 81 + 243/15 - 0)
= 216π / 5
but that includes the "cylinder" from x = 1 to 2
which is π(2-1)^2 = π
final volume = 216π/5 - π = 211π/5 cubic units
check my arithmetic
To find the volume of the solid obtained by rotating the region bounded by the curves about the specified line, you can use the method of cylindrical shells. The volume V can be calculated by the following integral:
V = ∫[a, b] 2πr(x)h(x) dx
where a and b are the x-values that define the region, r(x) is the distance between the axis of rotation (x = 2) and the curve at x, and h(x) is the height of the shell at x.
In this case, the region is bounded by the curves y = 27x^3, y = 0, x = 1, and we are rotating it about x = 2.
First, let's find the limits of integration. The region is bounded on the left by x = 1, so our lower limit of integration will be a = 1. The region is unbounded on the right, so we need to find the x-value where the curves intersect. Setting y = 27x^3 and y = 0 equal to each other, we can solve for x:
27x^3 = 0
x = 0
So the upper limit of integration will be b = 0.
Next, let's calculate r(x) and h(x). Since the axis of rotation is x = 2, the distance between x and 2 is given by r(x) = 2 - x. The height is given by h(x) = y = 27x^3.
Now we can set up the integral:
V = ∫[1, 0] 2π(2 - x)(27x^3) dx
Simplifying this expression, we have:
V = 2π∫[1, 0] (54x^3 - 27x^4) dx
Integrating term by term, we get:
V = 2π [13.5x^4 - 5.4x^5] between x = 1 and x = 0
Evaluating this integral, we have:
V = 2π [(13.5(1)^4 - 5.4(1)^5) - (13.5(0)^4 - 5.4(0)^5)]
V = 2π (13.5 - 0)
V = 27π
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = 27x^3, y = 0, x = 1 about x = 2 is 27π cubic units.