Solución
Para calcular la masa del resorte, debemos encontrar el valor de la integral de línea escalar ∫ C ( e x + y z ) d s \int_C (e^x + yz)ds ∫ C ( e x + yz ) d s C C C t t t
∫ C ( e x + y z ) d s = ∫ 0 π / 2 ( ( e t + 4 c o s t s e n t ) 1 + ( − 2 c o s t ) 2 + ( 2 s e n t ) 2 ) d t = ∫ 0 π / 2 ( ( e t + 4 c o s t s e n t ) 5 ) d t = 5 [ e t + 2 s e n 2 t ] t = 0 t = π / 2 = 5 ( e π / 2 + 1 ) \begin{aligned}
\int_C (e^x + yz)ds &= \int_0^{\pi/2}\bigg((e^t + 4cos\;ts en\;t)\sqrt{1 + (−2 cos\; t)^2 + (2 sen\; t)^2}\bigg)dt\\
&= \int_0^{\pi/2}\big((e^t + 4cos\; t sen\;t)\sqrt{5}\big)dt\\
&= \sqrt{5}\bigg[e^t + 2sen^2t\bigg]_{t=0}^{t=\pi/2}\\
&= \sqrt{5\big(e^{\pi/2}+1\big)}
\end{aligned} ∫ C ( e x + yz ) d s = ∫ 0 π /2 ( ( e t + 4 cos t se n t ) 1 + ( − 2 cos t ) 2 + ( 2 se n t ) 2 ) d t = ∫ 0 π /2 ( ( e t + 4 cos t se n t ) 5 ) d t = 5 [ e t + 2 se n 2 t ] t = 0 t = π /2 = 5 ( e π /2 + 1 ) Por lo tanto, la masa es 5 ( e π / 2 + 1 ) k g \sqrt{5\big(e^{\pi/2}+1\big)}\;kg 5 ( e π /2 + 1 ) k g