Answer is: option3
\( \dfrac{49}{6} \pi \)Solution:
Find the points of intersection:
- Intersection of \( y = \sqrt{x + 1} \) and \( y = x - 1 \):
\[ \sqrt{x + 1} = x - 1 \]
Square both sides:
\[ x + 1 = (x - 1)^2 \Rightarrow x + 1 = x^2 - 2x + 1 \]
Simplify:
\[ x^2 - 3x = 0 \Rightarrow x(x - 3) = 0 \Rightarrow x = 0 \text{ or } x = 3 \]
- At \( x = 0 \): \( y = \sqrt{0 + 1} = 1 \), and \( y = 0 - 1 = -1 \), so they do not intersect at \( x = 0 \)
- The valid intersection is at \( x = 3 \), where \( y = 2 \)
Intersection with the y-axis (\( x = 0 \)):
- For \( y = \sqrt{x + 1} \): When \( x = 0 \), \( y = 1 \)
- For \( y = x - 1 \): When \( x = 0 \), \( y = -1 \)
Bounds for \( x \): from \( x = 0 \) to \( x = 3 \)
When rotating about \( y = 2 \), the volume \( V \) is given by:
\[ V = \pi \int_a^b \left( [R_{\text{outer}}(x)]^2 - [R_{\text{inner}}(x)]^2 \right) dx \]
Where:
- \( R_{\text{outer}}(x) = 2 - \text{(lower function)} = 2 - (x - 1) = 3 - x \)
- \( R_{\text{inner}}(x) = 2 - \text{(upper function)} = 2 - \sqrt{x + 1} \)
- Bounds: \( x = 0 \) to \( x = 3 \)
Volume Integral:
\[ V = \pi \int_0^3 \left[ (3 - x)^2 - (2 - \sqrt{x + 1})^2 \right] dx \]
Expand the integrand:
\[ (3 - x)^2 = 9 - 6x + x^2 \]
\[ (2 - \sqrt{x + 1})^2 = 4 - 4\sqrt{x + 1} + x + 1 = 5 + x - 4\sqrt{x + 1} \]
Subtract the two:
\[ (9 - 6x + x^2) - (5 + x - 4\sqrt{x + 1}) = 4 - 7x + x^2 + 4\sqrt{x + 1} \]
Now integrate term by term:
\[ V = \pi \int_0^3 \left( 4 - 7x + x^2 + 4\sqrt{x + 1} \right) dx \]
\[ = \pi \left[ 4x - \frac{7}{2}x^2 + \frac{1}{3}x^3 + 4 \cdot \frac{2}{3}(x + 1)^{3/2} \right]_0^3 \]
\[ = \pi \left[ 4x - \frac{7}{2}x^2 + \frac{1}{3}x^3 + \frac{8}{3}(x + 1)^{3/2} \right]_0^3 \]
\[ = \pi \left( \frac{49}{6} \right) \]
Option C