Chapter 14.5, Problem 18E

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347

Chapter
Section

### Calculus (MindTap Course List)

11th Edition
Ron Larson + 1 other
ISBN: 9781337275347
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)=16âˆ’x2âˆ’y2.

Formula Used:

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

S=âˆ¬R1+[fx(x,y)]2+[fy(x,y)]2dA

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

The differentiation formula is ddx(xn)=nxnâˆ’1,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)=16âˆ’x2âˆ’y2.

Now, on performing partial differentiation on it with respect to x, make sure to use the formula, ddx(xn)=nxnâˆ’1,ddx(constant)=0.

fx(x,y)=ddx(16âˆ’x2âˆ’y2)=âˆ’2x2âˆ’1+0+0=âˆ’2x

Now, with respect to y.

The following can be found,

fy(x,y)=ddy(16âˆ’x2âˆ’y2)=0âˆ’2y2âˆ’1+0=âˆ’2y

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

S=âˆ¬R1+[2x]2+[âˆ’2y]2dA=âˆ¬R1+4x2+4y2dA=âˆ¬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=rcosÎ¸ and y=rsinÎ¸. 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 0â‰¤râ‰¤4 and 0â‰¤Î¸â‰¤Ï€4

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

We obtain the following,

S=âˆ«0Ï€2âˆ«041+4r2rdrdÎ¸

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

We get rdr=du8.

Perform integration on âˆ«041+4r2rdr first

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