Problem 1
1. Calculate \(\displaystyle\int\!\!\int\!\!\int_B 3x+y+z^2\, dV\) where \(B\) is the solid parallelepiped spanned by the vectors \(\mathbf{v}_1 = (1,1,1)\), \(\mathbf{v}_2 = (1,2,1)\), \(\mathbf{v}_3 = (2,2,1)\). (5 points)
Solution. Note that the corresponding transformation from the unit cube to the given parallelepiped is \(F(u,v,w) = (u+v+2w,\, u+2v+2w,\, u+v+w)\). The Jacobian of this transformation is given by
\[ J_F = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 2 & 2 & 1 \end{vmatrix} = |-1| = 1. \]
We perform the change of variables to get \[ \begin{align*} \int\!\!\int\!\!\int_B 3x+y+z^2\, dV &= \int_0^1\!\int_0^1\!\int_0^1 \left[3(u+v+2w) + (u+2v+2w) + (u+v+w)^2\right]\cdot J_F\, du\, dv\, dw \\ &= 1\cdot\int_0^1\!\int_0^1\!\int_0^1 3u+3v+6w+u+2v+2w+u^2+v^2+w^2+2uv+2vw+2uw\, du\, dv\, dw \\ &= \int_0^1\!\int_0^1\!\int_0^1 4u+u^2+2uv+2uw+5v+v^2+8w+w^2+2vw\, du\, dv\, dw \\ &= \int_0^1\!\int_0^1 \frac{7}{3}+6v+v^2+2vw+9w+w^2\, dv\, dw \\ &= \int_0^1 \frac{7}{3}+3+\frac{1}{3}+10w+w^2\, dw = \frac{8}{3}+3+5+\frac{1}{3} = 11. \end{align*} \]
Problem 2
2. Find the unit tangent vector for \(\mathbf{r}(t) = (6,\, \cos(3t),\, 3\sin(4t))\) for \(0\le t\le 2\pi\). (5 points)
Solution. A tangent vector \(\mathbf{r}'(t)\) is given by \(\mathbf{r}'(t) = (0,\, -3\sin(3t),\, 12\cos(4t))\). We find the magnitude
\[ ||\mathbf{r}'(t)|| = \sqrt{0^2+(-3\sin(3t))^2+(12\cos(4t))^2} = \sqrt{9\sin^2(3t)+144\cos^2(4t)}. \]
Combining we get the unit tangent vector \[ \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \left(0,\ \frac{-3\sin(3t)}{\sqrt{9\sin^2(3t)+144\cos^2(4t)}},\ \frac{12\cos(4t)}{\sqrt{9\sin^2(3t)+144\cos^2(4t)}}\right). \]