Which of the following integrals correctly computes the volume formed when the region is bounded by the curves x^2+y^2=25 x=3 and y=0

irene003   ·   10.01.2020 10:31
ANSWER(S)
answered: keykey5349
29.06.2019 22:00

he has 140 chickens because he has 10 times more chickens than ducks and he has 14 ducks so 14*10=140

answered: williamsjako
29.06.2019 16:10

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.

answered: parosil
28.06.2019 12:00

part a: 1.08

part b: 27.59

answered: cesarcastellan9
11.10.2020 06:28

The correct integral is  V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy  ⇒ 2nd answer

Step-by-step explanation:

The formula of the volume of a curve y = f(x) by integration around the y-axis is V=\pi \int\limits^b_a {[f(y)]^{2} } \, dy  

∵ The curves are x² + y² = 25 , x = 3 and y = 0

- Lets find x in terms of y in the first curve

∵ x² + y² = 25

- Subtract y² from both sides

∴ x² = 25 - y²

- Take square root for both sides

∴ x = \sqrt{25-y^{2}}

To find the limit of the integration b substitute x by 3

∵ x = 3

∵ x² + y² = 25

∴ 3² + y² = 25

∴ 9 + y² = 25

- Subtract 9 from both sides

∴ y² = 16

- Take square root for both sides

∴ y = ± 4

∵ The 1st limit a is at y = 0

∴ a = 0

∵ The 2nd limit is b at y = 4

∴ b = 4

∴ b = 4

There are two curves x = 3 and x = \sqrt{25-y^{2}}

To find the volume integrate the difference of the squares of the equations of the two curves and multiply the difference by π

∵  V=\pi \int\limits^b_a {[[f(y)]^{2}-[g(y)]^{2}}] \, dy  

- Subtract square 3 from the square of  \sqrt{25-y^{2}}    

∴   V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy    

The correct integral is    V=\pi \int\limits^4_0 {[(\sqrt{25-y^{2}})^{2}-3^{2} }] \, dy    

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