Question

Find the volume of the solid generated by revolving the area bounded by y=x^2 and x-axis from [0,2] around the x-axis

Answer 1

Answer:  $\dfrac{32\pi }{5}\ \text{unit}^3$

Step-by-step explanation:

Given

The area is bounded by $y=x^2$ and x-axis from [0,2]

The volume generated for $y=f(x)$ when rotated about the x-axis in the interval [a,b] is

$V=\int_a^b\pi y^2dx$

Insert the values

$\Rightarrow V=\int_0^2\pi x^4dx\\\\\Rightarrow V=\pi \int_0^2x^4dx\\\\\Rightarrow V=\pi \left[ \dfrac{x^5}{5}\right]_0^2\\\\\Rightarrow V=\dfrac{2^5\pi }{5}\\\\\Rightarrow V=\dfrac{32\pi }{5}\ \text{unit}^3$