Evaluate the line integral along the curve C.
-$\int _ { C } \left( \frac { 4 } { z } \right) ^ { 1 / 3 } \mathrm { ds } , \mathrm { C }$ is the curve $\mathbf { r } ( \mathrm { t } ) = \left( 4 \mathrm { t } ^ { 3 } \cos \mathrm { t } \right) \mathbf { i } + \left( 4 \mathrm { t } ^ { 3 } \sin \mathrm { t } \right) \mathbf { j } + 4 \mathrm { t } ^ { 3 } \mathbf { k } , 0 \leq \mathrm { t } \leq 3 \sqrt { 2 }$
A) $36 ( 4 - \sqrt { 2 } )$
B) $72 ( 4 + \sqrt { 2 } )$
C) 0
D) $72 ( 4 - \sqrt { 2 } )$

Question

Quizplus · Accepted Answer

Using the formula for line integrals, we have:

$$\int_C \left(\frac{4}{z}\right)^{1/3} ds = \int_a^b \left(\frac{4}{z(t)}\right)^{1/3} |\mathbf{r}'(t)| dt$$

where $z(t) = 4t^3$, $\mathbf{r}'(t) = (12t^2\cos t)\mathbf{i} + (12t^2\sin t)\mathbf{j} + (12t^2)\mathbf{k}$, and $|\mathbf{r}'(t)| = 12t^2$.

Plugging these into the integral, we get:

$$\int_0^{3\sqrt{2}} \left(\frac{4}{4t^3}\right)^{1/3} (12t^2) dt = \int_0^{3\sqrt{2}} 4\ dt = 12\sqrt{2}$$

Therefore, the correct answer is D) $72(4-\sqrt{2})$.

Evaluate the Line Integral Along the Curve C ∫C(4z)1/3ds,C\int _ { C } \left( \frac { 4 } { z } \right) ^ { 1 / 3 } \mathrm { ds } , \mathrm { C }∫C​(z4​)1/3ds,C

Evaluate the Line Integral Along the Curve C $\int _ { C } \left( \frac { 4 } { z } \right) ^ { 1 / 3 } \mathrm { ds } , \mathrm { C }$