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]

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