V = π ∫ from -1 to 1 (( (y - 1/2)² + (1 - y)² )) dy

Steps to Solve

1. Set Up the Integral

The volume ( V ) is given by the integral

$$ V = \pi \int_{-1}^{1} \left( \left( \frac{y - 1}{2} \right)^{2} + \left( \frac{1 - y}{2} \right)^{2} \right) dy. $$

2. Simplify Inside the Integral

Next, simplify the expression inside the integral.

First, calculate ( \left( \frac{y - 1}{2} \right)^{2} ) and ( \left( \frac{1 - y}{2} \right)^{2} ):

$$ \left( \frac{y - 1}{2} \right)^{2} = \frac{(y - 1)^{2}}{4} $$ $$ \left( \frac{1 - y}{2} \right)^{2} = \frac{(1 - y)^{2}}{4} $$

The combined expression becomes: $$ \frac{(y - 1)^{2}}{4} + \frac{(1 - y)^{2}}{4} = \frac{1}{4}((y - 1)^{2} + (1 - y)^{2}) $$

3. Calculate the Squared Terms

Next, expand ( (y - 1)^{2} + (1 - y)^{2} ):

$$ (y - 1)^{2} = y^{2} - 2y + 1 $$ $$ (1 - y)^{2} = y^{2} - 2y + 1 $$

Thus, $$ (y - 1)^{2} + (1 - y)^{2} = 2y^{2} - 4y + 2 $$

4. Integrate the Simplified Expression

Now replace the original integral with the simplified version:

$$ V = \pi \int_{-1}^{1} \frac{1}{4} (2y^{2} - 4y + 2) dy $$

This simplifies to:

$$ V = \frac{\pi}{4} \int_{-1}^{1} (2y^{2} - 4y + 2) dy $$

5. Break Down the Integration

Now, split the integral:

$$ V = \frac{\pi}{4} \left( \int_{-1}^{1} 2y^{2} dy - 4 \int_{-1}^{1} y dy + 2 \int_{-1}^{1} 1 dy \right) $$

6. Evaluate Each Integral

• For ( \int_{-1}^{1} 1 , dy ):

$$ \int_{-1}^{1} 1 , dy = 2 $$

• For ( \int_{-1}^{1} y , dy ):

$$ \int_{-1}^{1} y , dy = 0 $$

• For ( \int_{-1}^{1} 2y^{2} , dy ):

$$ \int_{-1}^{1} 2y^{2} , dy = 2 \cdot \left( \frac{y^{3}}{3} \Big|_{-1}^{1} \right) = 2 \cdot \left( \frac{1}{3} - \left(-\frac{1}{3}\right) \right) = \frac{4}{3} $$

Putting all of this back into the expression for ( V ):

$$ V = \frac{\pi}{4} \left( \frac{4}{3} - 0 + 2 \cdot 2 \right) = \frac{\pi}{4} \left( \frac{4}{3} + 4 \right) = \frac{\pi}{4} \left( \frac{4 + 12}{3} \right) = \frac{\pi}{4} \cdot \frac{16}{3} = \frac{4\pi}{3} $$