   Chapter 14.5, Problem 18E

Chapter
Section
Textbook Problem

Finding Surface Area In Exercises 17-20, find the area of the surface.The portion of the paraboloid z = 16 − x 2 − y 2 in the first octant

To determine

To calculate: The area of the surface given that is given in the question as f(x,y)=16x2y2 in the first octant of the given paraboloid.

Explanation

Given:

The region is denoted by f(x,y)=16x2y2.

Formula Used:

The surface area of the region R can be calculated by,

S=R1+[fx(x,y)]2+[fy(x,y)]2dA

Apply the power rule of integration, that is, xndx=xn+1n+1.

The differentiation formula is ddx(xn)=nxn1,ddx(constant)=0.

The equation of the circle is x2+y2=r2, where r represents the radius.

Calculation:

The given function is written as f(x,y)=16x2y2.

Now, on performing partial differentiation on it with respect to x, make sure to use the formula, ddx(xn)=nxn1,ddx(constant)=0.

fx(x,y)=ddx(16x2y2)=2x21+0+0=2x

Now, with respect to y.

The following can be found,

fy(x,y)=ddy(16x2y2)=02y21+0=2y

Substitute the calculated value in the formula S=R1+[fx(x,y)]2+[fy(x,y)]2dA.

S=R1+[2x]2+[2y]2dA=R1+4x2+4y2dA=R1+4x2+4y2dA

For the given region, the limits need to be converted to polar points in case of both x and y.

Then, the polar curve, x=rcosθ and y=rsinθ. This implies to the following,

x2+y2=r2(cos2x+sin2x)=r2=16

Since the region is in the first octant, the obtained region can be described as,

The region is, then, bounded by 0r4 and 0θπ4

Thus, on applying limits to the integral using the formula xndx=xn+1n+1,

We obtain the following,

S=0π2041+4r2rdrdθ

On substituting u=(1+4r2),du=8rdr.

We get rdr=du8.

Perform integration on 041+4r2rdr first

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