The portion of the parabola y=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface area generated isA=8/3a[(h+a)/-a/2]Use the result to find the value of h if the parabola y=36x when revolved about the x-axis is to have surface area 1000.

Mathematics College

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-axis. Show that the surface area generated is
A=8/3π√a[(h+a)³/²-a³/2]
Use the result to find the value of h if the parabola y²=36x when revolved about the x-axis is to have surface area 1000.

Answers

Answer 1

Answer:

See below for Part A.

Part B)

[tex]\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614[/tex]

Step-by-step explanation:

Part A)

The parabola given by the equation:

[tex]y^2=4ax[/tex]

From 0 to h is revolved about the x-axis.

We can take the principal square root of both sides to acquire our function:

[tex]y=f(x)=\sqrt{4ax}[/tex]

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

[tex]\displaystyle S=2\pi\int_{a}^{b}r(x)\sqrt{1+\big[f^\prime(x)]^2} \,dx[/tex]

Where r(x) is the distance between f and the axis of revolution.

From the sketch, we can see that the distance between f and the AoR is simply our equation y. Hence:

[tex]r(x)=y(x)=\sqrt{4ax}[/tex]

Now, we will need to find f’(x). We know that:

[tex]f(x)=\sqrt{4ax}[/tex]

Then by the chain rule, f’(x) is:

[tex]\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}[/tex]

For our limits of integration, we are going from 0 to h.

Hence, our integral becomes:

[tex]\displaystyle S=2\pi\int_{0}^{h}(\sqrt{4ax})\sqrt{1+\Big(\frac{2a}{\sqrt{4ax}}\Big)^2}\, dx[/tex]

Simplify:

[tex]\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax}\Big(\sqrt{1+\frac{4a^2}{4ax}}\Big)\,dx[/tex]

Combine roots;

[tex]\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax\Big(1+\frac{4a^2}{4ax}\Big)}\,dx[/tex]

Simplify:

[tex]\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx[/tex]

Integrate. We can consider using u-substitution. We will let:

[tex]u=4ax+4a^2\text{ then } du=4a\, dx[/tex]

We also need to change our limits of integration. So:

[tex]u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2[/tex]

Hence, our new integral is:

[tex]\displaystyle S=2\pi\int_{4a^2}^{4ah+4a^2}\sqrt{u}\, \Big(\frac{1}{4a}\Big)du[/tex]

Simplify and integrate:

[tex]\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big][/tex]

Simplify:

[tex]\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big][/tex]

FTC:

[tex]\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big][/tex]

Simplify each term. For the first term, we have:

[tex]\displaystyle (4ah+4a^2)^\frac{3}{2}[/tex]

We can factor out the 4a:

[tex]\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}[/tex]

Simplify:

[tex]\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}[/tex]

For the second term, we have:

[tex]\displaystyle (4a^2)^\frac{3}{2}[/tex]

Simplify:

[tex]\displaystyle =(2a)^3[/tex]

Hence:

[tex]\displaystyle =8a^3[/tex]

Thus, our equation becomes:

[tex]\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big][/tex]

We can factor out an 8a^(3/2). Hence:

[tex]\displaystyle S=\frac{\pi}{3a}(8a^\frac{3}{2})\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big][/tex]

Simplify:

[tex]\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big][/tex]

Hence, we have verified the surface area generated by the function.

Part B)

We have:

[tex]y^2=36x[/tex]

We can rewrite this as:

[tex]y^2=4(9)x[/tex]

Hence, a=9.

The surface area is 1000. So, S=1000.

Therefore, with our equation:

[tex]\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big][/tex]

We can write:

[tex]\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big][/tex]

Solve for h. Simplify:

[tex]\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big][/tex]

Divide both sides by 8π:

[tex]\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27[/tex]

Isolate term:

[tex]\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}[/tex]

Raise both sides to 2/3:

[tex]\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9[/tex]

Hence, the value of h is:

[tex]\displaystyle h=\Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}-9\approx7.4614[/tex]

The Portion Of The Parabola Y=4ax Above The X-axis, Where Is Form 0 To H Is Revolved About The X-axis.

Answer 2

You seem to have left out that 0 ≤ x ≤ h.

From y² = 4ax, we get that the top half of the parabola (the part that lies in the first quadrant above the x-axis) is given by y = √(4ax) = 2√(ax). Then the area of the surface obtained by revolving this curve between x = 0 and x = h about the x-axis is

[tex]2\pi\displaystyle\int_0^h y(x) \sqrt{1+\left(\frac{\mathrm dy(x)}{\mathrm dx}\right)^2}\,\mathrm dx[/tex]

We have

y(x) = 2√(ax) → y'(x) = 2 • a/(2√(ax)) = √(a/x)

so the integral is

[tex]4\sqrt a\pi\displaystyle\int_0^h \sqrt x \sqrt{1+\frac ax}\,\mathrm dx[/tex]

[tex]=\displaystyle4\sqrt a\pi\int_0^h (x+a)^{\frac12}\,\mathrm dx[/tex]

[tex]=4\sqrt a\pi\left[\dfrac23(x+a)^{\frac32}\right]_0^h[/tex]

[tex]=\dfrac{8\pi\sqrt a}3\left((h+a)^{\frac32}-a^{\frac32}\right)[/tex]

Now, if y² = 36x, then a = 9. So if the area is 1000, solve for h :

[tex]1000=8\pi\left((h+9)^{\frac32}-27\right)[/tex]

[tex]\dfrac{125}\pi=(h+9)^{\frac32}-27[/tex]

[tex]\dfrac{125+27\pi}\pi=(h+9)^{\frac32}[/tex]

[tex]\left(\dfrac{125+27\pi}\pi\right)^{\frac23}=h+9[/tex]

[tex]\boxed{h=\left(\dfrac{125+27\pi}\pi\right)^{\frac23}-9}[/tex]

Related Questions

[tex]\frac{d}{dx} \int t^2+1 \ dt[/tex]
There is a 2x on the bottom and x^2 on top of the integral symbol
Please help me my teacher did not teach us this:(

Answers

Answer:

[tex]\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} \ = \ 2x^5-8x^2+2x-2[/tex]

Step-by-step explanation:

[tex]\displaystyle{\frac{d}{dx} \int \limits_{2x}^{x^2} t^2+1 \ \text{dt} = \ ?[/tex]

We can use Part I of the Fundamental Theorem of Calculus:

[tex]\displaystyle\frac{d}{dx} \int\limits^x_a \text{f(t) dt = f(x)}[/tex]

Since we have two functions as the limits of integration, we can use one of the properties of integrals; the additivity rule.

The Additivity Rule for Integrals states that:

[tex]\displaystyle\int\limits^b_a \text{f(t) dt} + \int\limits^c_b \text{f(t) dt} = \int\limits^c_a \text{f(t) dt}[/tex]

We can use this backward and break the integral into two parts. We can use any number for "b", but I will use 0 since it tends to make calculations simpler.

[tex]\displaystyle \frac{d}{dx} \int\limits^0_{2x} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}[/tex]

We want the variable to be the top limit of integration, so we can use the Order of Integration Rule to rewrite this.

The Order of Integration Rule states that:

[tex]\displaystyle\int\limits^b_a \text{f(t) dt}\ = -\int\limits^a_b \text{f(t) dt}[/tex]

We can use this rule to our advantage by flipping the limits of integration on the first integral and adding a negative sign.

[tex]\displaystyle \frac{d}{dx} -\int\limits^{2x}_{0} t^2+1 \text{ dt} \ + \ \frac{d}{dx} \int\limits^{x^2}_0 t^2+1 \text{ dt}[/tex]

Now we can take the derivative of the integrals by using the Fundamental Theorem of Calculus.

When taking the derivative of an integral, we can follow this notation:

[tex]\displaystyle \frac{d}{dx} \int\limits^u_a \text{f(t) dt} = \text{f(u)} \cdot \frac{d}{dx} [u][/tex]where u represents any function other than a variable

For the first term, replace [tex]\text{t}[/tex] with [tex]2x[/tex], and apply the chain rule to the function. Do the same for the second term; replace

[tex]\displaystyle-[(2x)^2+1] \cdot (2) \ + \ [(x^2)^2 + 1] \cdot (2x)[/tex]

Simplify the expression by distributing [tex]2[/tex] and [tex]2x[/tex] inside their respective parentheses.

[tex][-(8x^2 +2)] + (2x^5 + 2x)[/tex] [tex]-8x^2 -2 + 2x^5 + 2x[/tex]

Rearrange the terms to be in order from the highest degree to the lowest degree.

[tex]\displaystyle2x^5-8x^2+2x-2[/tex]

This is the derivative of the given integral, and thus the solution to the problem.

Type the expression that results from the following series of steps:
Start with n and divide by 2.

Answers

the expression is n/2

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QUESTION GUIDES
3. An architect built a scale model of an office building using a scale in which
2.5 inches represents 20 feet. The height of the building is 240 feet.
12 in
What is the height of the scale model in inches?
30 in.
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80 in.
25 in.
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Answers

Answer:

30 inches

Step-by-step explanation:

240/20=12

12x2.5=30

Answer:

30 inches jrjdbdbrbejjr

Slope from graph 2
Find the Slope!

Answers

Answer:

y= -4x+4

Step-by-step explanation:

so the slope is -4x

Answer:

5/2 is the slope

Step-by-step explanation:

Rise/Run= 5/2

so, slope=5/2

The second derivative of a function F is given by F^11(x)=sin(3x)- cos (x^2). How many points of inflection does the graph FF have on the interval 0

Answers

Z+=^ plus deez d3+88 wayward too duke

Anita incorrectly wrote z =w-18 to describe the pattern in the table. Choose the best description of Anitas error

Answers

The answer choices are not present

Can someone please help me :(

Answers

Answer:

Step-by-step explanation:

2x + 1 = -3x + 6

add 3x

5x + 1 = 6

subtract 1

5x=5

Divide by 5

x=1

If this is correct, please mark brainliest!

Answer:

x=7

Step-by-step explanation:

2x + 1 = 15

-3x+6 = - 15

Since they are equal the equation is true

14. The area of a rectangle is 375 square inches and it’s perimeter is 80 inches. Find both the length and width.

Answers

Answer:

Length [tex]=25[/tex] inches

Width [tex]= 15[/tex] inches

Step-by-step explanation:

Let [tex]l[/tex] = length of rectangle, [tex]w[/tex] = width of rectangle

Given the area and the perimeter of the rectangle, we can write:

[tex]lw = 375[/tex]

[tex]2l+2w=80[/tex]

So:

[tex]l+w = 80\div2[/tex]

[tex]=40[/tex]

[tex]l=40-w[/tex]

Now, we can use substitution to find the value of [tex]w[/tex]:

[tex]lw = 375[/tex]

[tex](40-w)w = 375[/tex]

[tex]40w-w^2 = 375[/tex]

[tex]375+w^2-40w = 0[/tex] (Quadratic equation)

[tex](w-15)(w-25) = 0[/tex]

∴ [tex]w = 15, 25[/tex]

We can use substitution again to find the value of [tex]l[/tex]

[tex]lw = 375[/tex]

[tex]l=375\div w[/tex]

[tex]=25,15[/tex]

∵ Length usually refers to the longer side of a rectangle, ∴ length [tex]=25[/tex] inches and width [tex]=15[/tex] inches.

Hope this helps :)

The roots to a parabola are 2 and -3. What is one
possible equation for this parabola

Answers

Answer:

x^2+x-6

Step-by-step explanation:

i did a thing

hope this is correct and helps

Answer:

[tex]y = x^2 + x - 6[/tex]

Step-by-step explanation:

Hello!

We can utilize the factored form of a parabola to find a possible equation.

Factored Form: [tex]y = a(x - h)(x - k)[/tex]

h and k are roots

Since the roots of the parabola are given, we can plug in the roots for h and k, and plug in 1 for a and multiply to find the equation.

Find the Equation[tex]y = a(x - h)(x - k)[/tex][tex]y = 1(x - 2)(x +3)[/tex][tex]y = 1(x^2 + x - 6)[/tex][tex]y = x^2 + x - 6[/tex]

One possible equation is [tex]y = x^2 + x - 6[/tex].

What equation represents the proportional relationship displayed in the table ?

Answers

Answer:

y=5x

Step-by-step explanation:

Answer:

Step-by-step explanation:

When you divide each y-value by its corresponding x-value, you get 5.

For example, 10/2 = 5; 20/4 = 5

y is always 5 times x.

y = 5x

Answer: 5

If 3 times the difference of x and y is -30, and 2/3 of x plus 1/2 of y is 19, what are x and y

Answers

Answer:

x = 12 and y = 22

Step-by-step explanation:

It is given that,

3 times the difference of x and y is -30

3(x-y) = -30

3x-3y = -30 ....(1)

2/3 of x plus 1/2 of y is 19.

[tex]\dfrac{2x}{3}+\dfrac{y}{2}=19\\\\\dfrac{4x+3y}{6}=19\\\\4x+3y=114\ ...(2)[/tex]

Add equation (1) and (2).

3x-3y + 4x+3y = -30 +114

7x = 84

x = 12

Put the value of x in equation (1)

3x-3y = -30

3(12)-3y = -30

36+30 = 3y

3y = 66

y = 22

Hence, the value of x and y are 12 and 22.

dy÷dx=(x-1)(x+3) at x=2 first principal

Answers

Answer:

δy

δx = 2x − 4 + δx;

and the limit as δx → 0 is

dx = lim

δx→0

µδy

δx¶

= 2x − 4.

Step-by-step explanation:

dy/2d=(2-1)(2+3)

dy/2d=4+6-2-3

dy/2d=5

dy=5(2d)=10d

2d=5/dy

dy=5×(5/dy)= 25/dy

2d=5/dy

dy^2=25

dy=√25=5

2d=5/dy=5/5=1

d=1/2

dy=1/2×10=5 y=10

plug all of the values

5/(1/2×2)=5

4+6-2-3=5

so finally: 5=5

any questions?

Find the value of 7C7

Answers

Step-by-step explanation:

If there were 7 items, the number of ways to choose 7 out of the 7 items is only 1 way, which is to get all of them.

=> 7C7 = 1.

help again again again

Answers

Answer:

it's the last choice, by process of elimination that's the only one that includes the 4. and of course the math adds up

the answer is the last choice

Find sin θ. Right Triangle Trigonometry.

Answers

Answer:

A. 16/20

General Formulas and Concepts:

Trigonometry

[Right Triangles Only] SOHCAHTOA[Right Triangles Only] sin∅ = opposite over hypotenuse

Step-by-step explanation:

Step 1: Identify

Angle θ

Opposite Leg = 16

Hypotenuse = 20

Step 2: Find Ratio

Substitute [Sine]: sinθ = 16/20

Answer:

A. 16/20

Step-by-step explanation:

The sine ratio is the opposite leg over the hypotenuse, opp/hyp.

For angle theta, the opposite leg is 16. The hypotenuse is 20.

sin (theta) = 16/20

Answer: A. 16/20

Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.
Degree of Certainty
credits earned very uncertain Somewhat Certain very certain Row tot
under 10 24 16 6 46
10 through 59 16 8 20 44
60 or more 2 14 22 38
col total 42 38 48 128
Under the assumption of independence, the expected frequency in the upper left cell is:__________.
A. 2.47
B. 14.56
C. 2.00
D. 11.09

Answers

This question is incomplete, the complete question is;

Students in an introductory college economics class were asked how many credits they had earned in college, and how certain they were about their choice of major. Their replies are summarized below.

credits earned very deg.of certainty very Row Total

uncertain somewhat certain certain

under10 24 16 6 46

10 through 59 16 8 20 44

60 or more 2 14 22 38

col total 42 38 48 128

Under the assumption of independence, the expected frequency in the upper left cell is:__________.

A. 12.47

B. 14.56

C. 2.00

D. 11.09

Answer:

Under the assumption of independence, the expected frequency in the upper left cell is 12.47

Option A) 12.47 is the correct Option

Step-by-step explanation:

Given the data in table in the question;

the expected frequency in the upper left cell will be;

Eij = (Ti × Tj) / N

Row total Ti = 38

Column total Tj = 42

Total frequency N = 128

so we substitute

Eij = (38 × 42) / 128

= 1596 / 128

= 12.468 ≈ 12.47

Therefore Under the assumption of independence, the expected frequency in the upper left cell is 12.47

Option A) 12.47 is the correct Option

Evaluate 32 divided by 1.6
show work long division