Find the equation for the level surface of the function through the given point.
-$$f ( x , y , z ) = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t , \left( 4 , e ^ { 3 } , e ^ { - 5 } \right)$$
A) $- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } + 1 = \ln \theta + 1 + t e ^ { t }$
B) $- 6 e ^ { - 5 } - 2 e ^ { 3 } + 3 e ^ { 4 } + 1 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t$
C) $- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } - 7 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t$
D) $- 5 e ^ { - 5 } - 3 e ^ { 3 } + 3 e ^ { 4 } + 1 = \int _ { y } ^ { z } ( \ln \theta + 1 ) d \theta + \int _ { 0 } ^ { x } t e ^ { t } d t$

Question

Quizplus · Accepted Answer

We need to find the equation of the level surface passing through the point $(4, e^3, e^{-5})$.

For a level surface $f(x,y,z) = 	ext{constant}$, we can write the equation as
$$\int_y^z (\ln 	heta + 1) d	heta + \int_0^x te^t dt = C$$
where $C$ is the constant value of the level surface.

Plugging in the values of the point $(4, e^3, e^{-5})$, we get
$$\int_{e^{-5}}^{e^3} (\ln 	heta + 1) d	heta + \int_0^4 te^t dt = C$$
Simplifying the integrals, we get
$$\left[(\ln 	heta + 1)	hetaight]_{	heta=e^{-5}}^{	heta=e^3} + \left[(t-1)e^tight]_{t=0}^{t=4} = C$$
which gives
$$3e^4 - 5e^{-5} + 63 = C$$
Therefore, the equation of the level surface passing through $(4, e^3, e^{-5})$ is
$$\int_y^z (\ln 	heta + 1) d	heta + \int_0^x te^t dt = 3e^4 - 5e^{-5} + 63$$
Simplifying and comparing the options, we see that only option D matches this equation.

Find the Equation for the Level Surface of the Function