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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.

40. The part of the plane with vector equation r(uv) = ⟨u + v, 2 –3u, 1 + u v⟩ that is given by 0 ⩽ u ⩽ 2, −1 ⩽ v ⩽ 1

To determine

To find: The area of the part of the plane with vector equation r(u,v)=u+v,23u,1+uv,0u2,1v1 .

Explanation

Given data:

The vector equation of the part of the required plane is given as follows.

r(u,v)=u+v,23u,1+uv,0u2,1v1

Formula used:

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

A(S)=D|ru×rv|dA (1)

Here,

ru is the derivative of vector equation r(u,v) with respect to the parameter u and

rv is the derivative of vector equation r(u,v) with respect to the parameter v .

Write the expression to find ru .

ru=u[r(u,v)] (2)

Write the expression to find rv .

rv=v[r(u,v)] (3)

Calculation of ru :

Substitute u+v,23u,1+uv for r(u,v) in equation (2),

ru=uu+v,23u,1+uv=u(u+v),u(23u),u(1+uv)=(1+0),(03),(0+10)=1,3,1

Calculation of rv :

Substitute u+v,23u,1+uv for r(u,v) in equation (3),

rv=vu+v,23u,1+uv=v(u+v),v(23u),v(1+uv)=(0+1),(00),(0+01)=1,0,1

Calculation of ru×rv :

Substitute 1,3,1 for ru and 1,0,1 for rv in the expression ru×rv ,

ru×rv=1,3,1×1,0,1

Rewrite and compute the expression as follows

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Chapter 16 Solutions

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