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Thepotemich [5.8K]
2 years ago
14

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 ar

ea 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.​
Mathematics
2 answers:
castortr0y [4]2 years ago
8 0

Answer:

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 <em>h</em> 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 <em>r(x)</em> is the distance between <em>f</em> and the axis of revolution.

From the sketch, we can see that the distance between <em>f</em> and the AoR is simply our equation <em>y</em>. 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 <em>h</em>.

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

Nookie1986 [14]2 years ago
6 0

You seem to have left out that 0 ≤ <em>x</em> ≤ <em>h</em>.

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

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

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 <em>y</em>² = 36<em>x</em>, then <em>a</em> = 9. So if the area is 1000, solve for <em>h</em> :

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}

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