How to find inner surface area using integrals

Jaybird157

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Hello! One of the practice problems for my upcoming test was this problem. I think I have a pretty good understanding of how to calculate the surface areScreen Shot 2021-05-27 at 11.09.51 AM.pnga with integrals, but this problem completely threw me off. I only had two possible ideas, but I honestly have no idea where to go from here. Also, I asked my teacher and he just told me to make the segment between (-70, -16) linear. I didn't really know where to go from there.
I have an idea! If (-70,-16) is a line, you could break the dam wall into two parts; an upper and a lower. the upper part would be a perfect cylinder, so you wouldn't need to make an integral. I'm still stuck on the lower part. I'm honestly just really lost, but I'm trying to learn! I'd really appreciate any help.
 
Hello! One of the practice problems for my upcoming test was this problem. I think I have a pretty good understanding of how to calculate the surface areView attachment 27492a with integrals, but this problem completely threw me off. I only had two possible ideas, but I honestly have no idea where to go from here. Also, I asked my teacher and he just told me to make the segment between (-70, -16) linear. I didn't really know where to go from there.
I have an idea! If (-70,-16) is a line, you could break the dam wall into two parts; an upper and a lower. the upper part would be a perfect cylinder, so you wouldn't need to make an integral. I'm still stuck on the lower part. I'm honestly just really lost, but I'm trying to learn! I'd really appreciate any help.

The bottom part would be (part of) a frustum of a cone, so you wouldn't really need an integral there, either. But I presume you do.

What we need to see is, what do you know about finding surface areas by integration. Can you show an attempt, or at least a general formula?
 
This integral can help you.

[MATH]\frac{5\pi}{6}\int x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \ dx[/MATH]
Show us your thoughts, and we will give you a beautiful solution.
Do you have any idea what does this integral mean?
Do you know how to use it?

If you can use it properly, it can give you the slant inner surface area of the dam. The upper part of the dam is just the lateral surface area of a cylinder.
 
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