Question

Set up an integral of the form

dc h2(y) h1(y) f(x,y) dx dy
for the volume of the region enclosed by the surfaces z = 2 - x2 - y2 and z = 2x + 2y. What is h2(0)+f(0,0)?
A. 5.
B. 13 + 2.
C. V2+1.
D. V3+1.
E. V2 + 2.

Answer 1

Option D.

`\mathbf{√(3)} +1}`

Explanation:

The given integral is:

`\int^d_c \int ^(h_2(y))_(h_1(y)) f(x,y)\ dx \ dy`

The intersection curves enclosed by the surfaces:

z = 2 - x² - y² and z = 2x + 2y

This implies that:

`2x +2y = 2 - x^2- y^2`

`(x^2 + 2x) +(y^2 + 2y) = 2 \\ \\ (x+1)^2 + (y + 1)^2 = 4 --- (1)`

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)

`(x + 1)^2 = 4 - (y + 1)^2`

x + 1 = ±
`√(4 - (y + 1)^2)`

`x = -1 \pm √(4 - (y+1)^2)`

Now,

`h_2 (y) = -1 + √(4 - (y +1)^2)` and
`h_1 (y) = -1 - √(4 - (y +1)^2)`

`h_2(0) = -1 + √(4-1)`

`h_2(0) = -1 + √(3)`, and:

`At \ \ (-1, -1); \\ \\ z = 2 - (-1)^2 - (-1)^2 \\ \\ z = 2 - 1 - 1 \\ \\ z = 0`

and

`z = 2x + 2y \\ \\ z = 2(-1) + 2 (-1) \\ \\ z = -2 + (-2) \\ \\ z = -4`

The surface z = 2-x²-y² lies above z = 2x + 2y in the region of intersection

∴

`f(x,y) = (2 - x^2 - y^2) - (2x + 2y) \\ \\ f(0,0) = 2 - 0 \\ \\ f(0,0) = 2`

So, h₂ (0) + f(0,0) =
`√(3)} - 1 + 2`

h₂ (0) + f(0,0) =
`\mathbf{√(3)} +1}`