# Thread: calculate mass of the wire

1. ## calculate mass of the wire

Hello

If anyone is able to help me with this problem, I would greatly appreciate it. Since this is also an even numbered question, im not able to look up the answer. Special thanks to Scott who helped me with the Stoke's Theorem problem.

Question:

A wire is formed to the shape of the curve r(t) = 2 t i + 4/3 t^(3/2) j + 1/2 t^(2) k , for 0<= t <= 1. The density of the wire at the point (x,y,z) is given by p (x,y,z) = 4 x z + 9 y^(2) grams per linear unit. Calculate the mass of the wire.

Hello

If anyone is able to help me with this problem, I would greatly appreciate it. Since this is also an even numbered question, im not able to look up the answer. Special thanks to Scott who helped me with the Stoke's Theorem problem.

Question:

A wire is formed to the shape of the curve r(t) = 2 t i + 4/3 t^(3/2) j + 1/2 t^(2) k , for 0<= t <= 1. The density of the wire at the point (x,y,z) is given by p (x,y,z) = 4 x z + 9 y^(2) grams per linear unit. Calculate the mass of the wire.
You're given the parameterization of the curve: $x=2t$, $y=\tfrac{4}{3}t^{3/2}$ and $z=\tfrac{1}{2}t^2$. Note that $\left|\left|r^{\prime}(t)\right|\right|=\sqrt{4+4t +t^2}=t+2$

Thus, the density function becomes $\rho(t)=4(2t)\left(\tfrac{1}{2}t^{2}\right)+9\left (\tfrac{4}{3}t^{3/2}\right)^2=4t^3+16t^3=20t^3$.

Thus, the mass of the wire is defined by $\int_0^120t^3(t+2)\,dt$.

I'm sure you can take it from here.

3. ## Thanks Chris :)

Thanks Chris you ROCK!!!!!!!!! My attempt was waaaaaayyyyy off

Originally Posted by Chris L T521
You're given the parameterization of the curve: $x=2t$, $y=\tfrac{4}{3}t^{3/2}$ and $z=\tfrac{1}{2}t^2$. Note that $\left|\left|r^{\prime}(t)\right|\right|=\sqrt{4+4t +t^2}=t+2$

Thus, the density function becomes $\rho(t)=4(2t)\left(\tfrac{1}{2}t^{2}\right)+9\left (\tfrac{4}{3}t^{3/2}\right)^2=4t^3+16t^3=20t^3$.

Thus, the mass of the wire is defined by $\int_0^120t^3(t+2)\,dt$.

I'm sure you can take it from here.
Hi Chris,

I just wanna make sure I did it correctly, perhaps you can verify I have the correct answer. I got 14 linear units as the answer.