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3 years ago

The area of the entire figure is 1 square unit

Mathematics

1 answer:

Anton [14]3 years ago

4 0

Answer:

Area of shaded part = 0.42 unit²

Step-by-step explanation:

Given:

Length of shaded part = 0.7 unit

Width of shaded part = 3 x 2 = 0.6 unit

Find:

Area of shaded part

Computation:

Area of rectangle = L x B

Area of shaded part = 0.7 x 0.6

Area of shaded part = 0.42 unit²

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Use a surface integral to ﬁnd the surface area of the portion of the sphere xUse a surface integral to ﬁnd the surface area of t

brilliants [131]

Parameterize this surface (call it $S$ ) by

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

with $0\le u\le2\pi$ and $0\le v\le\dfrac\pi4$ . The limits on $u$ should be obvious. We find the upper limit for $v$ by solving for $v$ where the sphere and cone intersect:

$\begin{cases}x^2+y^2+z^2=1\\z=\sqrt{x^2+y^2}\end{cases}\implies x^2+y^2=\dfrac12$

$\implies(\cos u\sin v)^2+(\sin u\sin v)^2=\dfrac12$

$\implies\sin^2v=\dfrac12$

$\implies\sin v=\dfrac1{\sqrt2}\implies v=\dfrac\pi4$

Take the normal vector to $S$ to be

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

(orientation does not matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^{\pi/4}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv$

$=\displaystyle2\pi\int_0^{\pi/4}\sin v\,\mathrm dv=\boxed{(2-\sqrt2)\pi}$

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3 years ago

What is 4:24 in its simplest form?? :-)

Hitman42 [59]

I believe it is 1:6.

I hoped this helped :)

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3 years ago

Read 2 more answers

If x = -2 and y = 3, by how much does 2y2 - x exceed 2x2 - y?

Akimi4 [234]

Yes it does, to find this insert the values of x and y into the equation.
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4 years ago

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hichkok12 [17]

18.........
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