Find the volume of the region bounded above by the elliptical paraboloid z = 10 + x² + 3y² and below by the rectangle R: 0 ≤ x ≤ 1, 0 ≤ y ≤ 2.

Steps to Solve

1. Identify the Volume Integral

To find the volume under the elliptical paraboloid ( z = 10 + x^2 + 3y^2 ) and above the rectangle defined by ( R: 0 \leq x \leq 1 ) and ( 0 \leq y \leq 2 ), we'll set up a double integral:

$$ V = \iint_R z , dA $$

2. Set Up the Double Integral

Substituting the function ( z ) into the integral:

$$ V = \int_{0}^{1} \int_{0}^{2} (10 + x^2 + 3y^2) , dy , dx $$

Here, ( dA = dy , dx ).

3. Integrate with Respect to ( y )

First integrate the inner integral with respect to ( y ):

$$ \int_{0}^{2} (10 + x^2 + 3y^2) , dy $$

Calculating the integral:

• The antiderivative of ( 10 ) is ( 10y ).
• The antiderivative of ( x^2 ) is ( x^2y ).
• The antiderivative of ( 3y^2 ) is ( y^3 ).

Evaluating from ( 0 ) to ( 2 ):

$$ \int_{0}^{2} (10 + x^2 + 3y^2) , dy = [10y + x^2y + y^3]_{0}^{2} = [20 + 2x^2 + 24] - [0] = 20 + 2x^2 + 24 = 2x^2 + 44 $$

4. Integrate with Respect to ( x )

Now integrate the result with respect to ( x ):

$$ V = \int_{0}^{1} (2x^2 + 44) , dx $$

Calculating the integral:

• The antiderivative of ( 2x^2 ) is ( \frac{2}{3}x^3 ).
• The antiderivative of ( 44 ) is ( 44x ).

Evaluating from ( 0 ) to ( 1 ):

$$ \left[\frac{2}{3} x^3 + 44x\right]_{0}^{1} = \left(\frac{2}{3} + 44\right) - [0] = \frac{2}{3} + 44 = \frac{2}{3} + \frac{132}{3} = \frac{134}{3} $$

5. Final Volume Calculation

Thus, the volume ( V ) is:

$$ V = \frac{134}{3} $$