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Marizza181 [45]
3 years ago
13

Please help me 4and 5

Mathematics
1 answer:
hoa [83]3 years ago
5 0
What grade are you in

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G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
Please helppp! I don’t know the answer
disa [49]

Answer:

250x+140=y

Step-by-step explanation:

so amount earned for 10 hours of work is 250

8 0
3 years ago
PLEASE HELP<br><br> Need this ASAP
rosijanka [135]
I'm not sure but I think it is
4.48×10 10 <---this to is an exponent
4 0
3 years ago
Read 2 more answers
WILL MARK BRAINLIEST <br> in detail, what would you do in order to graph y = 3x -5
Hunter-Best [27]

So first, you need to identify the parts of this equation.

3x is the slope of the line, which is \frac{3}{1}.

-5 is the y-intercept. or the part of the line that touches the y-axis.

The first thing you want to do is plot the y-intercept. SO, go to the y-axis and find -5. Plot your point there.

Now, use the slope \frac{3}{1} to find your next plot. Since slope is \frac{rise}{run}, you go up three, and over to the right one.

Now, draw a line connecting the dots. Your slope should be positive and look like > /

~theLocoCoco

5 0
3 years ago
Which equation could have been used to create this graph?
Galina-37 [17]
D

Remember starts at zero slope is up 4 right 1
Y= 4/1x+0
Y=4x
8 0
3 years ago
Read 2 more answers
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