I need help asap i will give brainlist

Write the integral in one variable to find the volume of the solid obtained by rotating the first-quadrant region bounded by y =

Dovator [93]

Answer:

V = π ∫₀² (y² − 8y + 6√(2y)) dy

or

V = π ∫₀² (6x − 5x² + x³) dx

Step-by-step explanation:

y₁ = 0.5x²

y₂ = x

First, find the intersections of the curves.

0.5x² = x

x² = 2x

x² − 2x = 0

x (x − 2) = 0

x = 0 or x = 2

So the points of intersection are (0, 0) and (2, 2).

When we revolve this region about the line x = 3, we get a hollow shape that looks like an upside-down funnel, or a volcano.

One option is to use washer method to find the volume, by cutting a thin horizontal slice of thickness dy, inner radius 3−x₁ = 3−√(2y), and outer radius of 3−x₂ = 3−y.

V = ∫₀² π [(3−y)² − (3−√(2y))²] dy

V = ∫₀² π (9 − 6y + y² − 9 + 6√(2y) − 2y) dy

V = π ∫₀² (y² − 8y + 6√(2y)) dy

Another option is to use shell method to find the volume, by cutting a thin vertical slice of thickness dx, radius 3−x, and height y₂−y₁ = x−0.5x².

V = ∫₀² 2π (3 − x) (x − 0.5x²) dx

V = ∫₀² 2π (3x − 1.5x² − x² + 0.5x³) dx

V = ∫₀² 2π (3x − 2.5x² + 0.5x³) dx

V = π ∫₀² (6x − 5x² + x³) dx

The second option is arguably easier to evaluate, but either one will get you the same answer (V = 8π/3).

7 0

4 years ago

Find the area of a triangle with vertices at D(2, 0), E(–1, 0), and F(2, –7).

kati45 [8]

Im suck on the same question

8 0

3 years ago

Read 2 more answers

Fkhjglirugaeuzikjbvea HELP

kherson [118]

Step-by-step explanation: Isolate the variable by dividing each side by factors that don't contain the variable.

6 0

3 years ago

Find the volume of a right circular cone that has a height of 9 m and a base with a

AlladinOne [14]

Answer:

that is the solution to the question

3 0

4 years ago

(10 points) Consider the initial value problem y′+3y=9t,y(0)=7. Take the Laplace transform of both sides of the given differenti

Rashid [163]

Answer:

The solution

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3 t}$

Step-by-step explanation:

Explanation:-

Consider the initial value problem y′+3 y=9 t,y(0)=7

Step(i):-

Given differential problem

y′+3 y=9 t

Take the Laplace transform of both sides of the differential equation

L( y′+3 y) = L(9 t)

Using Formula Transform of derivatives

L(y¹(t)) = s y⁻(s)-y(0)

By using Laplace transform formula

$L(t) = \frac{1}{S^{2} }$

Step(ii):-

Given

L( y′(t)) + 3 L (y(t)) = 9 L( t)

$s y^{-} (s) - y(0) + 3y^{-}(s) = \frac{9}{s^{2} }$

$s y^{-} (s) - 7 + 3y^{-}(s) = \frac{9}{s^{2} }$

Taking common y⁻(s) and simplification, we get

$( s + 3)y^{-}(s) = \frac{9}{s^{2} }+7$

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

Step(iii):-

By using partial fractions , we get

$\frac{9}{s^{2} (s+3} = \frac{A}{s} + \frac{B}{s^{2} } + \frac{C}{s+3}$

$\frac{9}{s^{2} (s+3} = \frac{As(s+3)+B(s+3)+Cs^{2} }{s^{2} (s+3)}$

On simplification we get

9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Put s =0 in equation(i)

9 = B(0+3)

B = 9/3 = 3

Put s = -3 in equation(i)

9 = C(-3)²

C = 1

Given Equation 9 = A s(s+3) +B(s+3) +C(s²) ...(i)

Comparing 'S²' coefficient on both sides, we get

9 = A s²+3 A s +B(s)+3 B +C(s²)

0 = A + C

put C=1 , becomes A = -1

$\frac{9}{s^{2} (s+3} = \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}$

Step(iv):-

$y^{-}(s) = \frac{9}{s^{2} (s+3}+\frac{7}{s+3}$

$y^{-}(s) =9( \frac{-1}{s} + \frac{3}{s^{2} } + \frac{1}{s+3}) + \frac{7}{s+3}$

Applying inverse Laplace transform on both sides

$L^{-1} (y^{-}(s) ) =L^{-1} (9( \frac{-1}{s}) + L^{-1} (\frac{3}{s^{2} }) + L^{-1} (\frac{1}{s+3}) )+ L^{-1} (\frac{7}{s+3})$

By using inverse Laplace transform

$L^{-1} (\frac{1}{s} ) =1$

$L^{-1} (\frac{1}{s^{2} } ) = \frac{t}{1!}$

$L^{-1} (\frac{1}{s+a} ) =e^{-at}$

Final answer:-

Now the solution , we get

$Y (s) = 9( -1 +3 t + e^{-3 t} ) + 7 e ^{-3t}$

5 0

3 years ago