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3 years ago

What is the exact area and arc length of these sectors? Please helppp

Mathematics

1 answer:

Karolina [17]3 years ago

4 0

<u>Answers with step-by-step explanation:</u>

1. Area of sector 1 = $\frac{90}{360} \times \pi \times 12^2 = 36\pi$

2. Area of sector 2 = $\frac{45}{360} \times \pi \times 19^2 = \frac{2527}{8} \pi$

3. Area of sector 3 = $\frac{270}{360} \times \pi \times 15^2 = \frac{675}{4} \pi$

4. Area of sector 4 = $\frac{270}{360} \times \pi \times 6^2 = 27 \pi$

5. Arc length of sector 1 = $\frac{90}{360} \times 2 \times \pi \times 12 = 6\pi$

6. Arc length of sector 2 = $\frac{315}{360} \times 2 \times \pi \times 19 = \frac{133}{4} \pi$

7. Arc length of sector 3 = $\frac{270}{360} \times 2 \times \pi \times 15 = \frac{45}{2}\pi$

8. Arc length of sector 4 = $\frac{270}{360} \times 2 \times \pi \times 6 = 9\pi$

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2 years ago

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3 years ago

If y varies directly as x and y = 36 when x = 4,what is the value of y when x=6?

zepelin [54]

Answer:

Step-by-step explanation:

If the two numbers vary directly, we can use a proportion to find the missing value.

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2 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

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3 years ago