Questions: Consider the region R between y=x^2+3, x=0, x=2, and y=0. Use the Shell Method or Disk Method to find the volume V of the solid formed by rotation R around the y-axis. (Express numbers in exact form. Use symbolic notation and fractions where needed.) V=

Solution Steps

To find the volume of the solid formed by rotating the region \( R \) around the \( y \)-axis, we can use the Shell Method. The Shell Method involves integrating cylindrical shells. For a function \( y = f(x) \), the volume \( V \) is given by the integral:

\[ V = 2\pi \int_{a}^{b} x \cdot (f(x) - g(x)) \, dx \]

where \( f(x) \) is the outer function and \( g(x) \) is the inner function. In this case, \( f(x) = x^2 + 3 \) and \( g(x) = 0 \), with limits of integration from \( x = 0 \) to \( x = 2 \).

We are given the region \( R \) bounded by the curves \( y = x^2 + 3 \), \( x = 0 \), \( x = 2 \), and \( y = 0 \). The outer function is \( f(x) = x^2 + 3 \) and the inner function is \( g(x) = 0 \).

To find the volume \( V \) of the solid formed by rotating the region \( R \) around the \( y \)-axis using the Shell Method, we set up the integral:

\[ V = 2\pi \int_{0}^{2} x \cdot (f(x) - g(x)) \, dx = 2\pi \int_{0}^{2} x \cdot (x^2 + 3) \, dx \]

We compute the integral:

\[ V = 2\pi \int_{0}^{2} (x^3 + 3x) \, dx \]

Calculating the integral:

\[ \int (x^3 + 3x) \, dx = \frac{x^4}{4} + \frac{3x^2}{2} \]

Evaluating from \( 0 \) to \( 2 \):

\[ \left[ \frac{(2)^4}{4} + \frac{3(2)^2}{2} \right] - \left[ \frac{(0)^4}{4} + \frac{3(0)^2}{2} \right] = \left[ \frac{16}{4} + \frac{3 \cdot 4}{2} \right] = 4 + 6 = 10 \]

Thus, the volume is:

\[ V = 2\pi \cdot 10 = 20\pi \]

Final Answer