Find the volume of the region bounded above by the paraboloid z = x^2 + y^2 and below by the triangle enclosed by the lines y = x, x = 0, and x + y = 2 in the xy-plane.

Steps to Solve

1. Identify the bounds of the triangle in the xy-plane

The triangle is formed by the lines:

• ( y = x )
• ( x = 0 )
• ( x + y = 2 )

The vertices of the triangle can be found by solving these equations:

• Intersection of ( y = x ) and ( x + y = 2 ):

  • Substituting ( y = x ) into ( x + y = 2 ) gives ( x + x = 2 ), or ( x = 1 ). Thus, ( y = 1 ).
  • Vertex: ( (1, 1) ).

• The other intersections are:

  • Between ( x = 0 ) and ( x + y = 2 ): (0, 2).
  • Between ( x = 0 ) and ( y = x ): (0, 0).

The vertices of the triangle are ( (0, 0) ), ( (1, 1) ), and ( (0, 2) ).

2. Set up the double integral for volume

The volume ( V ) can be determined using the double integral: $$ V = \iint_{R} (z_{top} - z_{bottom}) , dA $$ where ( z_{top} = x^2 + y^2 ) (the paraboloid), and ( z_{bottom} = 0 ) (the triangle in the xy-plane).

3. Determine the region R for the double integral

We will integrate over the triangular region defined by the vertices. Choose ( y ) bounds from ( 0 ) to ( 2 - x ) for ( x ) ranging from ( 0 ) to ( 1 ).

4. Write the double integral

The volume becomes: $$ V = \int_{0}^{1} \int_{0}^{2-x} (x^2 + y^2) , dy , dx $$

5. Compute the inner integral

First, compute the inner integral: $$ \int_{0}^{2-x} (x^2 + y^2) , dy $$ Splitting it gives: $$ \int_{0}^{2-x} x^2 , dy + \int_{0}^{2-x} y^2 , dy $$ Calculating these:

• ( = x^2 (2-x) )
• ( = \left[ \frac{y^3}{3} \right]_{0}^{2-x} = \frac{(2-x)^3}{3} )

So we have: $$ \int_{0}^{2-x} (x^2 + y^2) , dy = x^2 (2-x) + \frac{(2-x)^3}{3} $$

6. Compute the outer integral

Now evaluate: $$ V = \int_{0}^{1} \left[ x^2 (2-x) + \frac{(2-x)^3}{3} \right] dx $$ This can be solved by expanding and integrating term by term.