Answer:
Option D.
[tex]\mathbf{\sqrt{3}} +1}[/tex]
Step-by-step explanation:
The given integral is:
[tex]\int^d_c \int ^{h_2(y)}_{h_1(y)} f(x,y)\ dx \ dy[/tex]
The intersection curves enclosed by the surfaces:
z = 2 - x² - y² and z = 2x + 2y
This implies that:
[tex]2x +2y = 2 - x^2- y^2[/tex]
[tex](x^2 + 2x) +(y^2 + 2y) = 2 \\ \\ (x+1)^2 + (y + 1)^2 = 4 --- (1)[/tex]
We will realize that the curve of this intersection is a circle which is centered at (-1, -1) of the radius 2.
So, from equation (1)
[tex](x + 1)^2 = 4 - (y + 1)^2[/tex]
x + 1 = ±[tex]\sqrt{4 - (y + 1)^2}[/tex]
[tex]x = -1 \pm \sqrt{4 - (y+1)^2}[/tex]
Now,
[tex]h_2 (y) = -1 + \sqrt{4 - (y +1)^2}[/tex] and [tex]h_1 (y) = -1 - \sqrt{4 - (y +1)^2}[/tex]
[tex]h_2(0) = -1 + \sqrt{4-1}[/tex]
[tex]h_2(0) = -1 + \sqrt{3}[/tex], and:
[tex]At \ \ (-1, -1); \\ \\ z = 2 - (-1)^2 - (-1)^2 \\ \\ z = 2 - 1 - 1 \\ \\ z = 0[/tex]
and
[tex]z = 2x + 2y \\ \\ z = 2(-1) + 2 (-1) \\ \\ z = -2 + (-2) \\ \\ z = -4[/tex]
The surface z = 2-x²-y² lies above z = 2x + 2y in the region of intersection
∴
[tex]f(x,y) = (2 - x^2 - y^2) - (2x + 2y) \\ \\ f(0,0) = 2 - 0 \\ \\ f(0,0) = 2[/tex]
So, h₂ (0) + f(0,0) = [tex]\sqrt{3}} - 1 + 2[/tex]
h₂ (0) + f(0,0) = [tex]\mathbf{\sqrt{3}} +1}[/tex]
