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3 years ago

Help please i dont understand

Mathematics

1 answer:

jenyasd209 [6]3 years ago

4 0

i cant see the entire question im sorry i dont understand what its asking

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Let C be the boundary of the region in the first quadrant bounded by the x-axis, a quarter-circle with radius 9, and the y-axis,

rewona [7]

Solution :

Along the edge $C_1$

The parametric equation for $C_1$ is given :

$x_1(t) = 9t , y_2(t) = 0 \ \ for \ \ 0 \leq t \leq 1$

Along edge $C_2$

The curve here is a quarter circle with the radius 9. Therefore, the parametric equation with the domain $0 \leq t \leq 1$ is then given by :

$x_2(t) = 9 \cos \left(\frac{\pi }{2}t\right)$

$y_2(t) = 9 \sin \left(\frac{\pi }{2}t\right)$

Along edge $C_3$

The parametric equation for $C_3$ is :

$x_1(t) = 0, \ \ \ y_2(t) = 9t \ \ \ for \ 0 \leq t \leq 1$

Now,

x = 9t, ⇒ dx = 9 dt

y = 0, ⇒ dy = 0

$\int_{C_{1}}y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

And

$x(t) = 9 \cos \left(\frac{\pi}{2}t\right) \Rightarrow dx = -\frac{7 \pi}{2} \sin \left(\frac{\pi}{2}t\right)$

$y(t) = 9 \sin \left(\frac{\pi}{2}t\right) \Rightarrow dy = -\frac{7 \pi}{2} \cos \left(\frac{\pi}{2}t\right)$

Then :

$\int_{C_1} y^2 x dx + x^2 y dy$

$=\int_0^1 \left[\left( 9 \sin \frac{\pi}{2}t\right)^2\left(9 \cos \frac{\pi}{2}t\right)\left(-\frac{7 \pi}{2} \sin \frac{\pi}{2}t dt\right) + \left( 9 \cos \frac{\pi}{2}t\right)^2\left(9 \sin \frac{\pi}{2}t\right)\left(\frac{7 \pi}{2} \cos \frac{\pi}{2}t dt\right) \right]$

$=\left[-9^4\ \frac{\cos^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} -9^4\ \frac{\sin^4\left(\frac{\pi}{2}t\right)}{\frac{\pi}{2}} \right]_0^1$

= 0

And

x = 0, ⇒ dx = 0

y = 9 t, ⇒ dy = 9 dt

$\int_{C_3} y^2 x dx + x^2 y dy = \int_0^1 (0)(0)+(0)(0) = 0$

Therefore,

$\oint y^2xdx +x^2ydy = \int_{C_1} y^2 x dx + x^2 x dx+ \int_{C_2} y^2 x dx + x^2 x dx+ \int_{C_3} y^2 x dx + x^2 x dx$

= 0 + 0 + 0

Applying the Green's theorem

$x^2 +y^2 = 81 \Rightarrow x \pm \sqrt{81-y^2}$

$\int_C P dx + Q dy = \int \int_R\left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dx dy$

Here,

$P(x,y) = y^2x \Rightarrow \frac{\partial P}{\partial y} = 2xy$

$Q(x,y) = x^2y \Rightarrow \frac{\partial Q}{\partial x} = 2xy$

$\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) = 2xy - 2xy = 0$

Therefore,

$\oint_Cy^2xdx+x^2ydy = \int_0^9 \int_0^{\sqrt{81-y^2}}0 \ dx dy$

$= \int_0^9 0\ dy = 0$

The vector field F is = $y^2 x \hat i+x^2 y \hat j$ is conservative.

5 0

3 years ago

What's is the answer

topjm [15]

Answer:

the answer

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4 0

3 years ago

Tate fills a container with 9 1/2 cups of lemonade. Tate drinks 32 fl. oz. of lemonade. How many fluid ounces are left?

iren [92.7K]

Answer:

44 fluid ounces of lemonade are left.

Step-by-step explanation:

Given:

Tate fills a container with 9 1/2 cups of lemonade.

Tate drinks 32 fl. oz. of lemonade.

Now, to find the quantity of fluid ounces left.

So, we convert the unit of cups into fluid ounces.

Tate fill the container of lemonade with = $9\frac{1}{2}\ cups.$

As, 1 cup = 8 fluid ounces.

So, by conversion factor:

$9\frac{1}{2} \times 8$

$=\frac{19}{2} \times 8$

$=9.5\times 8=76\ fl.\ oz.$

Thus, Tate fill the container of lemonade with = 76 fl. oz.

Tate drinks lemonade = 32 fl. oz.

Now, to get the fluid ounces left we subtract the quantity of lemonade Tate drink from the total quantity of lemonade filled in the container:

$76-32$

$=44\ fl.\ oz.$

Therefore, 44 fluid ounces of lemonade are left.

3 0

3 years ago

Solve: s = 4 + √ s + 2 a. s=2 b. s=7 c. s=2 or s=7 d. no real solution

lyudmila [28]

Rewrite s = 4 + sqrt(s+2) as s-4 = sqrt(s+2)

Square both sides: s^2 - 8s + 16 = s + 2

combine like terms: s^2 - 9s + 14 = 0

Factor this: (s-7)(s-2)=0

Solve for s: s={7, 2}

4 0

4 years ago

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The Elementary Functions (F18’s) are the set of functions of one real variable defined recursively as follows: Base cases: The i

Ira Lisetskai [31]

Answer:

sorry I don't know that

8 0

3 years ago