Find the area of the surface given by *z* = *f*(*x*, *y*) over the region *R*. (*Hint:* The integral may be simpler in polar coordinates.)

*f*(

*x*,

*y*) = 64 +

*x*^2 −

*y^*2

*R*= {(

*x*,

*y*):

*x*^2 +

*y^*2 ≤ 1}

### 1 Answer | Add Yours

We have the surface

`f(x,y) = 64 + x^2 - y^2`

and the constraint `x^2+y^2 <=1` , `x` and `y` `in R`

This constraint can be rewritten as

`x^2 = 1 - y^2` and `y^2 <=1`

Substituting for `x^2` , we want the area of the bounded region

`f(y) = 65 - 2y^2`

where `0 <= y <=1 `.

Integrating, the bounded region is

`B = int_0^1 (65 - 2y^2) = (65y - 2/3y^3)|_0^1 = 65 -2/3 = 64 1/3`

**The area is 64 + 1/3**

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes