Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=x^8/9 about the x-axis over the interval [1,2].
is that f(x) = x^(8/9) or f(x) = (x^8)/9
is will assume the latter.
Volume = π[integral) (x^16)/81 dx from 1 to 2
= π[x^17)/(17(81)) ] from 1 to 2
= π(2^17/(17(81)) - 1/(17(81))
= ....
Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=(2)/(x+1) about the x-axis over the interval [0,4].
Find the volume of the solid obtained by rotating the region under the graph of the function f(x)=e^x about the x-axis over the interval [0,2].
second one:
volume = π[integral] 4/(x+1)^2 from 0 to 2
= π((-4/(x+1)) from 0 to 2
= π(-4/3 - (-4/1)
= ....
third one:
you try it, the integral of (e^x)^2 is (1/2)e^(2x)
To find the volume of the solid obtained by rotating the region under the graph of the function f(x) = x^(8/9) about the x-axis over the interval [1,2], we can use the method of cylindrical shells.
The formula for calculating the volume using cylindrical shells is given by:
V = ∫[a,b] 2πx * f(x) * dx
where [a, b] is the interval along the x-axis, f(x) is the function, and dx represents the differential element.
In this case, we have:
V = ∫[1,2] 2πx * (x^(8/9)) * dx
To evaluate this integral, we can simplify the expression inside the integral:
V = 2π ∫[1,2] x^(17/9) * dx
Now we can integrate:
V = 2π * [(9/26) * x^(26/9)] |[1,2]
Substituting the limits of integration:
V = 2π * [(9/26) * (2^(26/9) - 1^(26/9))]
Simplifying further:
V = 2π * (9/26) * (2^(26/9) - 1)
This gives the volume of the solid obtained by rotating the region under the graph of f(x)=x^(8/9) about the x-axis over the interval [1,2].