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3 years ago

Find the volume of the given solid: Bounded by the cylinder y^2+z^2=4 and the planes x = 2y, x = 0, z = 0 in the first octant.

Mathematics

1 answer:

rodikova [14]3 years ago

3 0

The solution would be like this for this specific problem:

V = ∫ dV 
= ∫0→2 ∫ 0→π/2 ∫ 0→ 2·r·sin(φ) [ r ] dzdφdr 
= ∫0→2 ∫ 0→π/2 [ r·2·r·sin(φ) - r·0 ] dφdr 
= ∫0→2 ∫ 0→π/2 [ 2·r²·sin(φ) ] dφdr 
= ∫0→2 [ -2·r²·cos(π/2) + 2·r²·cos(0) ] dr 
= ∫0→2 [ 2·r² ] dr 
= (2/3)·2³ - (2/3)·0³ 
= 16/3

So the volume of the given solid is 16/3. I am hoping that these answers have satisfied your query and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

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The hypotenuse of the right triangle is 5y inches long. The lengths of the legs are x + 8 and x + 3 inches. If the perimeter of

Zolol [24]

Answer: 35 inches.

Step-by-step explanation:

We know that:

hypotenuse = 5*y in

cathetus 1 = (x + 8) in

cathetus 2 = (x + 3) in

The perimeter of the triangle is 76 inches, then:

5*y + (x + 8) + (x + 3) = 76

5*y + 2*x + 13 = 76

We also know that the length of the hypotenuse minus the length of the shorter leg is 17 in.

The shorter leg is x + 3, then:

5*y - (x + 3) = 17

Then we have the equations:

5*y + 2*x + 11 = 76

5*y - (x + 3) = 17

With only these two we can solve the system, first we need to isolate one of the variables in one of the equations, i will isolate x in the second equation.

x = 5*y - 3 - 17 = 5*y - 20

x = 5*y - 20

Now we can replace this in the other equation, we get:

5*y + 2*x + 11 = 76

5*y + 2*(5*y - 20) + 11 = 76

15*y - 40 + 13 = 76

15*y - 29 = 76

15*y = 76 + 29 = 105

and remember that the hypotenuse is equal to 5*y, then we want to get:

3*(5*y) = 105

5*y = 105/3 = 35

5*y = 35

Then te length of the hypotenuse is 35 inches.

6 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$