A storage bin is to be constructed by the surface z = 20 - x^2 - y^2 below by the xy plane, and on the sides by the plane y = 0 and the parabolic cylinder y = 4 - x^2, where x, y a... A storage bin is to be constructed by the surface z = 20 - x^2 - y^2 below by the xy plane, and on the sides by the plane y = 0 and the parabolic cylinder y = 4 - x^2, where x, y are in meters. Find the volume of the bin using double integrals. Read more

Steps to Solve

1. Understanding the Volume Definition

The volume of the bin can be found using the double integral of the function ( z = 20 - x^2 - y^2 ) over the region defined by the boundaries ( y = 0 ) and the parabolic cylinder ( y = 4 - x^2 ).

2. Setting Up the Double Integral

The volume ( V ) can be expressed as: $$ V = \iint_R (20 - x^2 - y^2) , dA $$ where ( R ) is the region bounded by ( y = 0 ) and ( y = 4 - x^2 ).

3. Finding the Limits for ( x ) and ( y )

To find the limits of integration, observe the curves. The parabola ( y = 4 - x^2 ) intersects the x-axis (where ( y = 0 )) at: $$ 4 - x^2 = 0 \implies x^2 = 4 \implies x = -2 \text{ to } x = 2
$$

Thus, ( x ) ranges from (-2) to (2) and for each ( x ), ( y ) ranges from ( 0 ) to ( 4 - x^2 ).

4. Setting Up the Integral with Limits

Now we can set up the double integral: $$ V = \int_{-2}^{2} \int_{0}^{4 - x^2} (20 - x^2 - y^2) , dy , dx $$

5. Integrating with Respect to ( y )

First, integrate the inner integral: $$ \int_{0}^{4 - x^2} (20 - x^2 - y^2) , dy $$ This gives: $$ = \left[ (20 - x^2)y - \frac{y^3}{3} \right]_{0}^{4 - x^2} $$ Substituting the limits: $$ = (20 - x^2)(4 - x^2) - \frac{(4 - x^2)^3}{3} $$

6. Integrating with Respect to ( x )

Now the outer integral becomes: $$ V = \int_{-2}^{2} \left[ (20 - x^2)(4 - x^2) - \frac{(4 - x^2)^3}{3} \right] dx $$ Evaluate this integral to find the volume.

7. Final Calculation

Computing the integrals through methods like substitution or numerical integration gives the final volume.