3.5: Practice Set

We can parameterize this straight line with \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle 1-t, -2t, 1-2t \right\rangle\) for \(0\le t\le 1\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle -1,-2,-2 \right\rangle\)

This derivative vector has length \(\|\vecs{r}'(t)\|=\sqrt{(-1)^2+(-2)^2+(-2)^2}=3\)

Therefore the line integral becomes:

\[\begin{align*} \int_C 4z\, ds &= \int_0^1 4z(t)\, \|\vecs{r}'(t)\|\,dt
\\[4pt] & = \int_0^1 4(1-2t)\cdot 3\, dt
\\[4pt] & = 12 \int_0^1 (1-2t) \, dt
\\[4pt] & = 12 \left[ t-t^2 \right] \Big|_0^1
\\[4pt] & = 12 \cdot 0
\\[4pt] & = 0 \end{align*} \]

We are given \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle t\cos(t)-\sin(t),t\sin(t)+\cos(t),t^2 \right\rangle\) for \(0\le t\le 2\pi \)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle -t\sin(t), t\cos(t), 2t \right\rangle\)

This derivative vector has length \(\|\vecs{r}'(t)\|=\sqrt{(-t\sin(t))^2+(t\cos(t))^2+(2t)^2}=\sqrt{5t^2}=t\sqrt{5}\)

Therefore the line integral becomes:

\[\begin{align*} \int_C e^z\, ds &= \int_0^{2\pi} e^{z(t)}\, \|\vecs{r}'(t)\|\,dt
\\[4pt] & = \int_0^{2\pi} e^{t^2}\cdot t\sqrt{5}\, dt
\\[4pt] & =\sqrt{5} \int_0^{2\pi} te^{t^2} \, dt
\\[4pt] & = \sqrt{5} \cdot \dfrac{1}{2}e^{t^2}  \Bigg|_0^{2\pi}
\\[4pt] & = \sqrt{5} \cdot \dfrac{e^{4\pi^2}-1}{2}
\\[4pt] & = \dfrac{(e^{4\pi^2}-1)\sqrt{5}}{2} \end{align*} \]

Let \(C_1\) be the line connecting \((1,2,2)\) to \((-1,-2,-2)\), and \(C_2\) the line connecting \((-1,-2,-2)\) to \((2,2,-2)\)

We can parameterize \(C_1\) with \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle 1-2t, 2-4t, 2-4t \right\rangle\) for \(0\le t\le 1\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle -2,-4,-4 \right\rangle\)

This derivative vector has length \(\|\vecs{r}'(t)\|=\sqrt{(-2)^2+(-4)^2+(-4)^2}=6\)

Therefore the line integral over \(C_1\) becomes:

\[\begin{align*} \int_{C_1} |x|\, ds &= \int_0^1 |x(t)|\, \|\vecs{r}'(t)\|\,dt
\\[4pt] & = \int_0^1 |1-2t| \cdot 6\, dt
\\[4pt] & = 6 \left( \int_0^{\frac{1}{2}} (1-2t) \, dt +\int_{\frac{1}{2}}^1 (2t-1) \, dt \right)
\\[4pt] & = 6\left( \left[ t-t^2 \right] \Big|_0^{\frac{1}{2}} +\left[ t^2-t \right] \Big|_{\frac{1}{2}}^1\right)
\\[4pt] & = 6\left( \dfrac{1}{4} + \dfrac{1}{4} \right)
\\[4pt] & = 3 \end{align*} \]

We can parameterize \(C_2\) with \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle -1+3t, -2+4t, -2 \right\rangle\) for \(0\le t\le 1\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle 3,4,0 \right\rangle\)

Which has length \(\|\vecs{r}'(t)\|=\sqrt{3^2+4^2+0^2}=5\)

Therefore the line integral over \(C_2\) becomes:

\[\begin{align*} \int_{C_2} |x|\, ds &= \int_0^1 |x(t)|\, \|\vecs{r}'(t)\|\,dt
\\[4pt] & = \int_0^1 |-1+3t| \cdot 5\, dt
\\[4pt] & = 5 \left( \int_0^{\frac{1}{3}} (1-3t) \, dt +\int_{\frac{1}{3}}^1 (3t-1) \, dt \right)
\\[4pt] & = 5\left( \left[ t-\dfrac{3}{2}t^2 \right] \Bigg|_0^{\frac{1}{3}} +\left[ \dfrac{3}{2}t^2-t \right] \Bigg|_{\frac{1}{3}}^1\right)
\\[4pt] & = 5\left( \dfrac{1}{6} + \dfrac{2}{3} \right)
\\[4pt] & = \dfrac{25}{6}\end{align*} \]

Adding these two together we get our final answer

\[ \int_{C} |x|\, ds= \int_{C_1} |x|\, ds+ \int_{C_2} |x|\, ds=3+\dfrac{25}{6}=\dfrac{43}{6} \nonumber \]

This curve \(C\) can be parameterized by \(\vecs{r}(t)=\langle x(t),y(t)\rangle=\left\langle \sin(t),\cos(t)\right\rangle\) for \(0\le t \le \dfrac{\pi}{2}\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle \cos(t),-\sin(t)\right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} x^2y\, dy &= \int_0^{\frac{\pi}{2}} [x(t)]^2\cdot y(t) \cdot y'(t)\,dt
\\[4pt] & = \int_0^{\frac{\pi}{2}} \sin^2(t)\cos(t)\cdot (-\sin(t)) \, dt
\\[4pt] & = \int_0^{\frac{\pi}{2}} -\sin^3(t)\cos(t) \, dt
\\[4pt] & =  -\dfrac{1}{4}\sin^4(t) \Bigg|_0^{\frac{\pi}{2}}
\\[4pt] & = -\dfrac{1}{4} \end{align*} \]

This curve \(C\) can be parameterized by \(\vecs{r}(t)=\langle x(t),y(t)\rangle=\left\langle \sqrt{\sin(t)},\sqrt{\cos(t)}\right\rangle\) for \(0\le t \le \dfrac{\pi}{2}\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle \dfrac{\cos(t)}{2\sqrt{\sin(t)}},-\dfrac{\sin(t)}{2\sqrt{\cos(t)}} \right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} x^2y\, dy &= \int_0^{\frac{\pi}{2}} [x(t)]^2\cdot y(t) \cdot y'(t)\,dt
\\[4pt] & = \int_0^{\frac{\pi}{2}} \sin(t)\sqrt{\cos(t)}\cdot \left(-\dfrac{\sin(t)}{2\sqrt{\cos(t)}}\right) \, dt
\\[4pt] & = \int_0^{\frac{\pi}{2}} -\dfrac{1}{2}\sin^2(t) \, dt
\\[4pt] & = \int_0^{\frac{\pi}{2}} \left(\dfrac{1}{4}\cos(2t)-\dfrac{1}{4}\right) \, dt
\\[4pt] & = \left[ \dfrac{1}{8}\sin(2t)-\dfrac{1}{4}t \right]\Bigg|_0^{\frac{\pi}{2}}
\\[4pt] & = -\dfrac{\pi}{8} \end{align*} \]

The curve \(C\) described in the intersection of a plane and paraboloid:

We can parameterize it using \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle -t, -t, 8-2t^2 \right\rangle\) for \(-2\le t\le 2\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle -1,-1,-4t \right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} (x+y)\, dz &= \int_{-2}^2 \left(x(t)+y(t)\right) \cdot z'(t)\,dt
\\[4pt] & = \int_{-2}^2 (-2t)\cdot (-4t) \, dt
\\[4pt] & = \int_{-2}^2 8t^2 \, dt
\\[4pt] & =  \dfrac{8}{3}t^3 \Bigg|_{-2}^2
\\[4pt] & = \dfrac{128}{3} \end{align*} \]

We can parameterize \(C\) with \( \vecs{r}(t)=\langle x(t),y(t)\rangle=\left\langle 1+t, 2+3t \right\rangle\) for \(0\le t\le 1\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle 1,3 \right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C}xy\,dx+y^2\,dy &= \int_0^1 \left( x(t)\cdot y(t) \cdot x'(t)+[y(t)]^2\cdot y'(t)\right) \,dt
\\[4pt] & = \int_0^1 \left( (1+t)(2+3t)\cdot 1+ (2+3t)^2\cdot 3\right)\, dt
\\[4pt] & = \int_0^1 (30t^2+41t+14) \, dt
\\[4pt] & = \left[ 10t^3+\dfrac{41}{2}t^2+14t \right]\Bigg|_0^1
\\[4pt] & = \dfrac{89}{2} \end{align*} \]

We are given \(\vecs{r}(t)=\langle t^2-2t+3,t^2+4t-7 \rangle \) over \(1\le t\le 3\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle 2t-2,2t+4 \right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C}(x-y)\, dx+(2x+y) \,dy &= \int_1^3 \left( (x(t)- y(t)) \cdot x'(t)+(2x(t)+y(t))\cdot y'(t)\right) \,dt
\\[4pt] & = \int_1^3 \left( ((t^2-2t+3)-(t^2+4t-7))\cdot (2t-2)+ (2(t^2-2t+3)+(t^2+4t-7 )) \cdot (2t+4)\right)\, dt
\\[4pt] & = \int_1^3 (6t^3+30t-24) \, dt
\\[4pt] & = \left[ \dfrac{3}{2}t^4+15t^2-24t \right]\Bigg|_1^3
\\[4pt] & = 192 \end{align*} \]

The curve \(C\) described in the intersection of a parabolic cylinder and a paraboloid:

We can parameterize it using \( \vecs{r}(t)=\langle x(t),y(t),z(t)\rangle=\left\langle t, t^2, 16-t^2-t^4 \right\rangle\) for \(-2\le t\le 2\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t),z'(t)\rangle=\left\langle 1,2t,-2t-4t^3 \right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} z\,dx+x \,dz &= \int_{-2}^2 \left(z(t)\cdot x'(t)+x(t) \cdot z'(t) \right) \,dt
\\[4pt] & = \int_{-2}^2 \left( (16-t^2-t^4)\cdot 1 + t\cdot(-2t-4t^3)\right) \, dt
\\[4pt] & = \int_{-2}^2 (-5t^4-3t^2+16) \, dt
\\[4pt] & = \left[-t^5-t^3+16t \right]\Big|_{-2}^2
\\[4pt] & = -16 \end{align*} \]

This curve \(C\) can be parameterized by \(\vecs{r}(t)=\langle x(t),y(t)\rangle=\left\langle \sin(t),\cos(t)\right\rangle\) for \(0\le t \le 2\pi\)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle \cos(t),-\sin(t)\right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} |x|\, dx+|y|\, dy &= \int_0^{2\pi} \left( |x(t)|\cdot x'(t)+|y(t)|\cdot y'(t)\right) \,dt
\\[4pt] & = \int_0^{2\pi} \left(|\sin(t)|\cdot \cos(t)+ |\cos(t)|\cdot (-\sin(t))\right) \,dt
\\[4pt] & = \left( \int_0^{\frac{\pi}{2}} 0 \, dt +\int_{\frac{\pi}{2}}^\pi 2\sin(t)\cos(t) \, dt + \int_{\pi}^{\frac{3\pi}{2}} 0 \, dt + \int_{\frac{3\pi}{2}}^{2\pi} -2\sin(t)\cos(t)\, dt \right)
\\[4pt] & = \left( \sin^2(t) \Big|_{\frac{\pi}{2}}^\pi+\left[ - \sin^2(t) \right]\Bigg|_{\frac{3\pi}{2}}^{2\pi} \right)
\\[4pt] & = \left( -1+1 \right)
\\[4pt] & = 0 \end{align*} \]

This curve \(C\) can be parameterized by \(\vecs{r}(t)=\langle x(t),y(t)\rangle=\left\langle \sin(t),\cos(t)\right\rangle\) for \(0\le t \le 2\pi \)

This gives derivative \(\vecs{r}'(t)=\langle x'(t),y'(t)\rangle=\left\langle \cos(t),-\sin(t)\right\rangle\)

Therefore the line integral over \(C\) becomes:

\[\begin{align*} \int_{C} |y|\, dx+|x|\, dy &= \int_0^{2\pi} \left( |y(t)|\cdot x'(t)+|x(t)|\cdot y'(t)\right) \,dt
\\[4pt] & = \int_0^{2\pi}\left( |\cos(t)|\cdot \cos(t)+|\sin(t)|\cdot (-\sin(t))\right) \,dt
\\[4pt] & = \left( \int_0^{\frac{\pi}{2}} \cos(2t) \, dt +\int_{\frac{\pi}{2}}^\pi -1 \, dt + \int_{\pi}^{\frac{3\pi}{2}} -\cos(2t) \, dt + \int_{\frac{3\pi}{2}}^{2\pi} 1\, dt \right)
\\[4pt] & = \left(  \dfrac{1}{2}\sin(2t) \Bigg|_0^{\frac{\pi}{2}}+\left[ -t \right]\Big|_{\frac{\pi}{2}}^\pi+\left[ -\dfrac{1}{2}\sin(2t) \right]\Bigg|_\pi^{\frac{3\pi}{2}}+  t \Big|_{\frac{3\pi}{2}}^{2\pi} \right)
\\[4pt] & = \left(0+\left(-\dfrac{\pi}{2}\right)+0+\dfrac{\pi}{2} \right)
\\[4pt] & = 0 \end{align*} \]