Answer:
See below for Part A. 
Part B)

Step-by-step explanation:
Part A)
The parabola given by the equation: 

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: 

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](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D2%5Cpi%5Cint_%7Ba%7D%5E%7Bb%7Dr%28x%29%5Csqrt%7B1%2B%5Cbig%5Bf%5E%5Cprime%28x%29%5D%5E2%7D%20%5C%2Cdx)
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: 

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

Then by the chain rule, f’(x) is: 

For our limits of integration, we are going from 0 to <em>h</em>.
Hence, our integral becomes: 

Simplify: 

Combine roots; 

Simplify: 

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

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

Hence, our new integral is: 

Simplify and integrate: 
![\displaystyle S=\frac{\pi}{2a}\Big[\,\frac{2}{3}u^{\frac{3}{2}}\Big|^{4ah+4a^2}_{4a^2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B%5Cpi%7D%7B2a%7D%5CBig%5B%5C%2C%5Cfrac%7B2%7D%7B3%7Du%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5CBig%7C%5E%7B4ah%2B4a%5E2%7D_%7B4a%5E2%7D%5CBig%5D)
Simplify: 
![\displaystyle S=\frac{\pi}{3a}\Big[\, u^\frac{3}{2}\Big|^{4ah+4a^2}_{4a^2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B%5Cpi%7D%7B3a%7D%5CBig%5B%5C%2C%20u%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%7C%5E%7B4ah%2B4a%5E2%7D_%7B4a%5E2%7D%5CBig%5D)
FTC: 
![\displaystyle S=\frac{\pi}{3a}\Big[(4ah+4a^2)^\frac{3}{2}-(4a^2)^\frac{3}{2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B%5Cpi%7D%7B3a%7D%5CBig%5B%284ah%2B4a%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D-%284a%5E2%29%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%5D)
Simplify each term. For the first term, we have: 

We can factor out the 4a: 

Simplify: 

For the second term, we have: 

Simplify: 

Hence: 

Thus, our equation becomes: 
![\displaystyle S=\frac{\pi}{3a}\Big[8a^\frac{3}{2}(h+a)^\frac{3}{2}-8a^3\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B%5Cpi%7D%7B3a%7D%5CBig%5B8a%5E%5Cfrac%7B3%7D%7B2%7D%28h%2Ba%29%5E%5Cfrac%7B3%7D%7B2%7D-8a%5E3%5CBig%5D)
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]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B%5Cpi%7D%7B3a%7D%288a%5E%5Cfrac%7B3%7D%7B2%7D%29%5CBig%5B%28h%2Ba%29%5E%5Cfrac%7B3%7D%7B2%7D-a%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%5D)
Simplify: 
![\displaystyle S=\frac{8\pi}{3}\sqrt{a}\Big[(h+a)^\frac{3}{2}-a^\frac{3}{2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B8%5Cpi%7D%7B3%7D%5Csqrt%7Ba%7D%5CBig%5B%28h%2Ba%29%5E%5Cfrac%7B3%7D%7B2%7D-a%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%5D)
Hence, we have verified the surface area generated by the function. 
Part B)
We have: 

We can rewrite this as: 

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]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20S%3D%5Cfrac%7B8%5Cpi%7D%7B3%7D%5Csqrt%7Ba%7D%5CBig%5B%28h%2Ba%29%5E%5Cfrac%7B3%7D%7B2%7D-a%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%5D)
We can write: 
![\displaystyle 1000=\frac{8\pi}{3}\sqrt{9}\Big[(h+9)^\frac{3}{2}-9^\frac{3}{2}\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%201000%3D%5Cfrac%7B8%5Cpi%7D%7B3%7D%5Csqrt%7B9%7D%5CBig%5B%28h%2B9%29%5E%5Cfrac%7B3%7D%7B2%7D-9%5E%5Cfrac%7B3%7D%7B2%7D%5CBig%5D)
Solve for h. Simplify: 
![\displaystyle 1000=8\pi\Big[(h+9)^\frac{3}{2}-27\Big]](https://tex.z-dn.net/?f=%5Cdisplaystyle%201000%3D8%5Cpi%5CBig%5B%28h%2B9%29%5E%5Cfrac%7B3%7D%7B2%7D-27%5CBig%5D)
Divide both sides by 8π: 

Isolate term: 

Raise both sides to 2/3: 

Hence, the value of h is: 
