Mathematics, 29.12.2020 14:00 ricky67

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: 3

Show answers

Answers

Answer from: datands

See below for Part A.

Part B)

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

Step-by-step explanation:

Part A)

The parabola given by the equation:

y^2=4ax

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

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

$y=f(x)=\sqrt{4ax}$

Please refer to the attachment below for the sketch.

The area of a surface of revolution is given by:

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

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:

$r(x)=y(x)=\sqrt{4ax}$

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

$f(x)=\sqrt{4ax}$

Then by the chain rule, f’(x) is:

$\displaystyle f^\prime(x)=\frac{1}{2\sqrt{4ax}}\cdot4a=\frac{2a}{\sqrt{4ax}}$

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

Hence, our integral becomes:

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

Simplify:

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

Combine roots;

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

Simplify:

$\displaystyle S=2\pi\int_{0}^{h}\sqrt{4ax+4a^2}\, dx$

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

$u=4ax+4a^2\text{ then } du=4a\, dx$

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

$u=4a(0)+4a^2=4a^2\text{ and } \\ u=4a(h)+4a^2=4ah+4a^2$

Hence, our new integral is:

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

Simplify and integrate:

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

Simplify:

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

FTC:

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

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

$\displaystyle (4ah+4a^2)^\frac{3}{2}$

We can factor out the 4a:

$\displaystyle =(4a)^\frac{3}{2}(h+a)^\frac{3}{2}$

Simplify:

$\displaystyle =8a^\frac{3}{2}(h+a)^\frac{3}{2}$

For the second term, we have:

$\displaystyle (4a^2)^\frac{3}{2}$

Simplify:

$\displaystyle =(2a)^3$

Hence:

$\displaystyle =8a^3$

Thus, our equation becomes:

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

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

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

Simplify:

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

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

Part B)

We have:

y^2=36x

We can rewrite this as:

y^2=4(9)x

Hence, a=9.

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

Therefore, with our equation:

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

We can write:

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

Solve for h. Simplify:

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

Divide both sides by 8π:

$\displaystyle \frac{125}{\pi}=(h+9)^\frac{3}{2}-27$

Isolate term:

$\displaystyle \frac{125}{\pi}+27=(h+9)^\frac{3}{2}$

Raise both sides to 2/3:

$\displaystyle \Big(\frac{125}{\pi}+27\Big)^\frac{2}{3}=h+9$

Hence, the value of h is:

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

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-ax

Answer from: shaunalab

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

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

We have

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

so the integral is

$4\sqrt a\pi\displaystyle\int_0^h \sqrt x \sqrt{1+\frac ax}\,\mathrm dx$

$=\displaystyle4\sqrt a\pi\int_0^h (x+a)^{\frac12}\,\mathrm dx$

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

$=\dfrac{8\pi\sqrt a}3\left((h+a)^{\frac32}-a^{\frac32}\right)$

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

$1000=8\pi\left((h+9)^{\frac32}-27\right)$

$\dfrac{125}\pi=(h+9)^{\frac32}-27$

$\dfrac{125+27\pi}\pi=(h+9)^{\frac32}$

$\left(\dfrac{125+27\pi}\pi\right)^{\frac23}=h+9$

$\boxed{h=\left(\dfrac{125+27\pi}\pi\right)^{\frac23}-9}$

Answer from: Quest

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing im pretty sure that is the right answer

step-by-step explanation:

Very urgent! for the figures below, assume they are made of semicircles, quarter circles and square

Answer from: Quest

answer: ay

fonsi

oh no, oh no

oh yeah

diridiri, dirididi daddy

sí, sabes que ya llevo un rato mirándote

tengo que bailar contigo hoy (dy)

vi que tu mirada ya estaba llamándome

muéstrame el camino que yo voy (oh)

tú, tú eres el imán y yo soy el metal

me voy acercando y voy armando el plan

solo con pensarlo se acelera el pulso (oh yeah)

ya, ya me está gustando más de lo normal

todos mis sentidos van pidiendo más

esto hay que tomarlo sin ningún apuro

despacito

quiero respirar tu cuello despacito

deja que te diga cosas al oído

para que te acuerdes si no estás conmigo

despacito

quiero desnudarte a besos despacito

firmo en las paredes de tu laberinto

y hacer de tu cuerpo todo un manuscrito (sube, sube, sube)

(sube, sube)

Another question on Mathematics

Mathematics, 21.06.2019 14:10

What is the perimeter, p, of a rectangle that has a length of x + 8 and a width of y − 1? p = 2x + 2y + 18 p = 2x + 2y + 14 p = x + y − 9 p = x + y + 7

Answers: 3

Answer

Mathematics, 21.06.2019 18:10

Television sizes are based on the length of the diagonal of the television's rectangular screen. so, for example, a 70-inch television is a television having a diagonal screen length of 70 inches. if a 70 inch television's diagonal forms a 29°angle with the base of the screen, what is the vertical height (1) of the screen to the nearest inch? 70 inhin

Answers: 3

Answer

Mathematics, 21.06.2019 20:00

Write the point-slope form of the line passing through (2, -12) and parallel to y=3x.

Answers: 2

Answer

Mathematics, 21.06.2019 21:00

M.xyz =radians. covert this radian measure to its equivalent measure in degrees.

Answers: 1

Answer

You know the right answer?

The portion of the parabola y²=4ax above the x-axis, where is form 0 to h is revolved about the x-ax...