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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the area of the surface.

47. The part of the paraboloid y = x2 + z2 that lies within the cylinder x2 + z2 = 16

To determine

To find: The area of part of the paraboloid y=x2+z2 that lies within the cylinder x2+z2=16 .

Explanation

Given data:

The equation of part of the paraboloid is given as follows.

y=x2+z2

The required part of the paraboloid lies within the cylinder x2+z2=16 .

Formula used:

Write the expression to find the surface area of the plane with the vector equation r(x,z) .

A(S)=D|rx×rz|dA (1)

Here,

rx is the derivative of vector equation r(x,z) with respect to the parameter x and

rz is the derivative of vector equation r(x,z) with respect to the parameter z .

Write the expression to find rx .

rx=x[r(x,z)] (2)

Write the expression to find rz .

rz=z[r(x,z)] (3)

Consider the x and z as parameters and parameterize the paraboloid as follows.

x=x,y=x2+z2,z=z

Write the vector equation of the paraboloid from the parametric equations as follows.

r(x,z)=xi+(x2+z2)j+zk

Calculation of rx :

Substitute xi+(x2+z2)j+zk for r(x,z) in equation (2),

rx=x[xi+(x2+z2)j+zk]=x(x)i+x(x2+z2)j+x(z)k=(1)i+(2x+0)j+(0)k=(1)i+2xj+(0)k

Calculation of rz :

Substitute xi+(x2+z2)j+zk for r(x,z) in equation (3),

rz=z[xi+(x2+z2)j+zk]=z(x)i+z(x2+z2)j+z(z)k=(0)i+(0+2z)j+(1)k=(0)i+2zj+(1)k

Calculation of rx×rz :

Substitute (1)i+2xj+(0)k for rx and (0)i+2zj+(1)k for rz in the expression rx×rz ,

rx×rz=[(1)i+2xj+(0)k]×[(0)i+2zj+(1)k]

Rewrite and compute the expression as follows.

rx×rz=|ijk12x002z1|=|2x02z1|i|1001|j+|12x02z|k=(2x0)i(10)j+(2z0)k=2xij+2zk

Substitute 2xij+2zk for rx×rz in equation (1),

A(S)=D|2xij+2zk|dA=D[(2x)2+(1)2+(2z)2]dA=D(4x2+1+4z2)dA

A(S)=D1+4(x2+z2)dA (4)

Consider the parametric equations for the cylinder x2+z2=16 as follows

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Chapter 16 Solutions

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