Mathematics

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.

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Answer

answered: 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

answered: 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

answered: Guest

19.

step-by-step explanation:

you add the 9 in the tens column since it is a single digit and 9 plus zero is 9 so it is 19

Answer

answered: Guest

$\omega=\{r3,r5,s3,s5,t3,t5\}$

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