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ohaa [14]
2 years ago
12

Please anwser fast and give an explanation step by step thank you

Mathematics
1 answer:
Ludmilka [50]2 years ago
7 0

What is the question?

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How do you find the volume of the solid generated by revolving the region bounded by the graphs
d1i1m1o1n [39]

Answer:

About the x axis

V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

Step-by-step explanation:

For this case we have the following functions:

y = 2x^2 , y=0, X=2

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on  y from 0 to 8.

We can find the area like this:

A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= 4\pi \int_{0}^2 x^4 dx

V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}

About the y axis

For this case we need to find the function in terms of x like this:

x^2 = \frac{y}{2}

x = \pm \sqrt{\frac{y}{2}} but on this case we are just interested on the + part x=\sqrt{\frac{y}{2}} as we can see on the second figure attached.

We can find the area like this:

A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy

V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)

And we can find the volume with this formula:

V = \int_{a}^b A(x) dx

V= \pi \int_{0}^2 64-32x^2 +4x^4 dx

V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}

And we can find the volume with this formula:

V = \int_{a}^b A(y) dy

V= \frac{\pi}{2} \int_{0}^8 y dy

V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi

6 0
3 years ago
How many 2 litre glasses of water are in a 3.6 litre pitcher
Lubov Fominskaja [6]
The answer would be '2'.
7 0
3 years ago
Over five different weeks, Irina tracked the hours she spent exercising and the hours she spent playing video games. What is the
madam [21]

Answer:

moderate negative correlation

6 0
3 years ago
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Solve the following system by any method. –8x – 8y = 0 –8x + 2y = –20
madam [21]
This one is simple since we already have the two x variables.equal. All we have to do is subtract the equations from one another to get the answer.
So i will subtract the left side by the other left side and the right side by the other right side

-8x - 8y -(-8x + 2y) = 0 -(-20)

distribute negative sign

-8x - 8y + 8x - 2y = 0 + 20

do the math
- 10y = 20
Y = -2

plug t into an equation
-8x -8 (-2) = 0

-8x + 16 = 0
-8x = -16
x = 2
answer (2, -2)
8 0
3 years ago
Solve for x. What is x/15 = 21/5?​
melisa1 [442]

Answer: x = 63

Step-by-step explanation:

15/3 = 5

3 * 21 = 63

63/15 = 21/5

8 0
3 years ago
Read 2 more answers
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