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RSB [31]
2 years ago
13

Suppose that surface σ is parameterized by r(u,v)=⟨ucos(3v),usin(3v),v⟩, 0≤u≤7 and 0≤v≤2π3 and f(x,y,z)=x2+y2+z2. Set up the sur

face integral (you don't need to evaluate it). ∫σ∫f(x,y,z)dS=∫R∫f(x(u,v),y(u,v),z(u,v))∥∥∥∂r∂u×∂r∂v∥∥∥dA = 2pi/3 ∫ 0 7 ∫ 0 u^2+v^2 ∥∥∥ ∥∥∥dudv
Mathematics
1 answer:
Bad White [126]2 years ago
6 0

Looks like you have most of the details already, but you're missing one crucial piece.

\sigma is parameterized by

\vec r(u,v)=\langle u\cos3v,u\sin3v,v\rangle

for 0\le u\le7 and 0\le v\le\frac{2\pi}3, and a normal vector to this surface is

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left\langle\sin3v,-\cos3v,3u\right\rangle

with norm

\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{\sin^23v+(-\cos3v)^2+(3u)^2}=\sqrt{9u^2+1}

So the integral of f(x,y,z)=x^2+y^2+z^2 is

\displaystyle\iint_\sigma f(x,y,z)\,\mathrm dA=\boxed{\int_0^{2\pi/3}\int_0^7(u^2+v^2)\sqrt{9u^2+1}\,\mathrm du\,\mathrm dv}

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3x + 5y = 7<br> How to solve it
Lubov Fominskaja [6]

Answer:

x=1 and y=4/5

Step-by-step explanation:

<u>Given system of equations</u>

3x+5y=7

5x+10y=13

<u />

<u>Multiply first equation by 2</u>

2(3x+5y)=2(7)

6x+10y=14

<u>Eliminate y-variable to get "x"</u>

 6x+10y=14

<u>- 5x+10y=13</u>

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<u>Substitute x=1 into either equation and find "y"</u>

3x+5y=7

3(1)+5y=7

3+5y=7

5y=4

y=4/5

Therefore, x=1 and y=4/5

6 0
2 years ago
★write the pair of integer in which multiplication is -45 and difference is 12.
vladimir1956 [14]

Answer:

(3 and 15)

Step-by-step explanation:

If the product of both of these two numbers needs to be a negative 45 then there are no pair of integers that comply with both requirements.

If the multiplication in the question is wrong and it is a positive 45 then the pair of integers that would comply with these requirements would be (3 and 15). Multiplying this pair together would give you 45 and the difference between them is also 12 meaning it complies with both requirements that were asked for in the question.

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2 years ago
Solve the system of equations
rewona [7]

Answer:

x = 1/4

y = -1/2

z = 9/4

Step-by-step explanation:

Here we have a system of 3 equations with 3 variables:

4*x + 2*y + 1 = 1

2*x - y = 1

x + 3*y + z = 1

The first step to solve this, is to isolate one of the variables in one of the equations, let's isolate "y" in the second equation:

2*x - y = 1

2*x - 1 = y

Now that we have an expression equivalent to "y", we can replace this in the other two equations:

4*x + 2*(2*x - 1) + 1 = 1

x + 3*(2*x - 1) + z = 1

Now let's simplify these two equations:

8*x - 1 = 1

7*x - 3 + z = 1

Now, in the first equation we have only the variable x, so we can solve that equation to find the value of x:

8*x - 1 = 1

8*x = 1 + 1 = 2

x = 2/8 = 1/4

Now that we know the value of x, we can replace this in the other equation to find the value of z.

7*(1/4) -3 + z = 1

7/4 - 3 + z = 1

z = 1 + 3 - 7/4

z = 4 - 7/4

z = 16/4 - 7/4 = 9/4

z = 9/4

Now we can use the equation y = 2*x - 1 and the value of x to find the value of y:

y = 2*(1/4) - 1

y = 2/4 - 1

y = 1/2 - 1

y = -1/2

Then the solution is:

x = 1/4

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z = 9/4

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