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Find the Equation for the Level Surface of the Function

Question 202

Multiple Choice

Find the equation for the level surface of the function through the given point.
- f(x,y,z) =yz(lnθ+1) dθ+0xtetdt,(4,e3,e5) f ( x , y , z ) = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t , \left( 4 , e ^ { 3 } , e ^ { - 5 } \right)


A) 5e53e3+3e4+1=lnθ+1+tet- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } + 1 = \ln \theta + 1 + t e ^ { t }
B) 6e52e3+3e4+1=yz(lnθ+1) dθ+0xtetdt- 6 e ^ { - 5 } - 2 e ^ { 3 } + 3 e ^ { 4 } + 1 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t
C) 5e53e3+3e47=yz(lnθ+1) dθ+0xtetdt- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } - 7 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t
D) 5e53e3+3e4+1=yz(lnθ+1) dθ+0xtetdt- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } + 1 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t

Correct Answer:

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