Evaluate $\int _ { C } y z d y + x y d z$, where $C$ is given by $x = 10 \sqrt { t } , y = 3 t , z = 10 t ^ { 2 } , 0 \leq t \leq 1$. Round your answer to two decimal place.
A) $194.93$
B) $193.93$
C) $195.93$
D) $196.93$
E) $197.93$

Question

Quizplus · Accepted Answer

To evaluate the given integral $\int _ { C } y z d y + x y d z$, we substitute $x = 10 \sqrt { t } , y = 3 t , z = 10 t ^ { 2 }$ and also express $dy$ and $dz$ in terms of $dt$. Given:- $x = 10\sqrt{t}$- $y = 3t$- $z = 10t^2$Differentiating $x, y, z$ with respect to $t$, we get:- $dx/dt = 5t^{-1/2}$- $dy/dt = 3$- $dz/dt = 20t$Substituting $x, y, z, dy, dz$ into the integral, we get:$$\int_{0}^{1} (3t)(10t^2)(3)dt + (10\sqrt{t})(3t)(20t)dt$$$$= \int_{0}^{1} (90t^3 + 600t^{5/2})dt$$$$= \left[ \frac{90t^4}{4} + \frac{600t^{7/2}}{7/2} \right]_0^1$$$$= \left[ \frac{90}{4} + \frac{1200}{7} \right]$$$$= 22.5 + 171.43$$$$= 193.93$$Therefore, the correct answer is $193.93$, which corresponds to choice B.

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