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Marrrta [24]
3 years ago
10

Evaluate the line integral in Stokes? Theorem to evaluate the surface integral integrate S ( x F) n dS. Assume that n is in the

positive z-direction. F=(x+y,y+z,z+x); Sis the tilted disk enclosed by r(t)=(cost,4 sin t root 5 cost). Integrate s ( * F). n ds= (Type an exact answer. using pi as needed)
Mathematics
1 answer:
gtnhenbr [62]3 years ago
6 0

Answer:

Step-by-step explanation:

Consider that F(x,y,z) = (x + y, y + z, z + z)

r(t)=(cos t,4sint,\sqt{5}cost)\\r'(t)=(-sint,4cost,-\sqrt{5}sint)\\F(r(t))=[tex]\int\limits^{2\pi}_0 {F(r(t))r'(t)} \, dt \\\\=\int\limits^{2\pi}_0(cost+4sint,4sint+\sqrt{5}cost+cost)(-sint,4cost,-\sqt{5}sint)dt\\\\=\int\limits^{2\pi}_0(-sintcost-4sin^2t+16costsint+4\sqrt{5}cos^2t-5sintcost-\sqrt{5}sintcost)dt\\\\=\int\limits^{2\pi}_0(10sintcost-4sin^2t+4\sqrt{5}cos^2t-\sqrt{5}sintcost)dt\\\\=\int\limits^{2\pi}_0((10-\sqrt{5})sintcost-(4+4+\sqrt{5})sin^2t+4\sqrt{5})dt\\\\(10-\qrt{5})(0)-(4+4+\sqrt{5}(\pi)+4\sqrt{5}(2\pi)\\\\(\sqrt{5}-1)4\pi)[/tex]

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How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y
son4ous [18]

Solution:

Given:

x^2=6y

Part A:

The vertex of an up-down facing parabola of the form;

\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}

Rewriting the equation given;

\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\  \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\  \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\  \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\  \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}

Therefore, the focus is;

(0,\frac{3}{2})

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

\begin{gathered} 4p(y-k)=(x-h)^2 \\  \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}

Rewriting the equation in standard form,

\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}

Therefore, the directrix is;

y=-\frac{3}{2}

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1 year ago
A figure is rotated and reflected. what can u say about the new figure in relation to the original figure?
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The new figure is formed differently and is not the same as the other ome
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3 years ago
A basketball gymnasium is 25 meters high, 80 meters wide and 200 meters long. We want to connect two strings, one from each of t
Rudiy27

Answer:

a) the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be 216.85 m

c) the angle that is formed by the strings is  1.23 rad

Step-by-step explanation:

assuming that we stand on one of the corners on the floor , so our coordinates are (0,0,0)  , then the coordinates of the center of the gymnasium   are found through

x center = (25 + 0)/2 = 12.5 m

y center = (80+ 0)/2 = 40 m

z center = (200+ 0)/2 = 100 m

then the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be the modulus of the vector that points from our position to the diagonally opposite corners

|S| = √(25²+80²+200²) = 216.85 m

c) the angle can be found through the dot product of the vectors that represent the strings S₁ and S₂

S₁ =(25,80,10)

S₂ =(-25,80,100)

then

S₁*S₂ = 25*(-25) +80*80 + 100*100 = 15775

but also

S₁*S₂ = |S₁||S₂| cos θ = |S|² * cos θ

S₁*S₂ =  |S|² * cos θ

cos θ= S₁*S₂/|S|²

θ= cos ⁻¹ ( S₁*S₂/|S|² ) = cos ⁻¹ [15775/(25²+80²+200²)] = 1.23 rad

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Chris walked the dog for 80 minutes or 1 hour and 20 minutes
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3 years ago
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