NEED HELP ASAP! SHORT QUESTION!

Find the volume of the solid generated when R (shaded region) is revolved about the given line. x=6−3sec y, x=6, y= π 3, and

Dmitrij [34]

Answer:

$V=9\pi\sqrt{3}$

Step-by-step explanation:

In order to solve this problem we must start by graphing the given function and finding the differential area we will use to set our integral up. (See attached picture).

The formula we will use for this problem is the following:

$V=\int\limits^b_a {\pi r^{2}} \, dy$

where:

$r=6-(6-3 sec(y))$

$r=3 sec(y)$

a=0

$b=\frac{\pi}{3}$

so the volume becomes:

$V=\int\limits^\frac{\pi}{3}_0 {\pi (3 sec(y))^{2}} \, dy$

This can be simplified to:

$V=\int\limits^\frac{\pi}{3}_0 {9\pi sec^{2}(y)} \, dy$

and the integral can be rewritten like this:

$V=9\pi\int\limits^\frac{\pi}{3}_0 {sec^{2}(y)} \, dy$

which is a standard integral so we solve it to:

$V=9\pi[tan y]\limits^\frac{\pi}{3}_0$

so we get:

$V=9\pi[tan \frac{\pi}{3} - tan 0]$

which yields:

$V=9\pi\sqrt{3}$ ]

6 0

4 years ago

Part 2: Solve the following word problem. Show all work to receive full credit..

lys-0071 [83]

The numerical representation and final elevation of the fish relative to sea level are :

Final position = -13.52 + 7.8
Final position = - 5.72

Given the Parameters :

Initial depth of fish = - 13.52

Change in elevation of fish = 7.8 feets

Positions below sea level are represented as negative values (-) ;

Hence, initial position of fish = - 13.52

Change in position = Rise in depth of 7.8 feets

The numerical representation of the final position of the fish can be written thus :

Final position = Initial position + change in position

Final position = - 13.52 + 7.8

Hence, the final position of the fish will be :

Final position of fish = -13.52 + 7.8 = - 5.72

Learn more : brainly.com/question/18109354

6 0

3 years ago

Read 2 more answers

nhập kho 5500 kg nguyên vật liệu A với giá nhập chưa có thế GTGT 10% là 52000đ/kg . đã thanh toán bằng chuyển khoảng

Julli [10]

Answer:

i dont know

Step-by-step explanation:

8 0

3 years ago

I don't need an explanation, I would just like the answer to these questions

avanturin [10]

1)HGI ,S.S.S

2)DEF,A.S.A

3)C.E.D,S.A.S

6 0

4 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago