Answer:
[tex]\dfrac{127}{7} \pi \textsf{or } 60 [/tex] cubic units
Step-by-step explanation:
To find the volume of the solid generated by rotating the curve [tex]y = x^3[/tex] from [tex]x = 1[/tex] to [tex]x = 2[/tex] around the y-axis, we can use the disk method.
The volume [tex]V[/tex] is given by the integral:
[tex] V = \pi \int_{a}^{b} [f(x)]^2 \, dx [/tex]
In this case, [tex]a = 1[/tex] and [tex]b = 2[/tex], and [tex]f(x) = x^3[/tex].
[tex] V = \pi \int_{1}^{2} x^6 \, dx [/tex]
Now, integrate with respect to [tex]x[/tex]:
[tex] V = \pi \left[ \dfrac{1}{7}x^7 \right]_{1}^{2} [/tex]
Evaluate the expression at [tex]x = 2[/tex] and subtract the value at [tex]x = 1[/tex]:
[tex] V = \pi \left( \dfrac{1}{7}(2^7) - \dfrac{1}{7}(1^7) \right) [/tex]
[tex] V = \pi \left( \dfrac{128}{7} - \dfrac{1}{7} \right) [/tex]
[tex] V = \pi \cdot \dfrac{127}{7} [/tex]
[tex] V = \dfrac{127}{7}\pi [/tex]
[tex] V \approx 56.99746672 [/tex]
[tex] V \approx 60 \textsf{ (in nearest whole number)}[/tex]
So, the volume of the solid generated is [tex]\dfrac{127}{7} \pi \textsf{or } 60 [/tex] cubic units.
