Cal 2 help please - hard problem

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The base of a particularly curvy building is described by $$y=x*sin(3x)+10$$ with $$y$$ in feet and $$x$$ is in tens of feet for $$0 \leq x \leq 3\pi$$. Views are shown in the attached file i uploaded. The cross sections perpendicular to the x-axis and paraell to the y-axis are semi-circular.

that was just the intro, here are the questions:
can someone just help me setup the integral to these three problems?

1.) Local health codes require that the air must completely filtered (or replaced) every half hour for the building. What must be the minimum capacity in cubic feet per minute to of the air handler (pump) to meet the code?
okay this question is kinda confusing. I'm thinking it's a surface area question.here's what i did:

dy/dx = 3x*cos(3x) + sin(3x)

surface area formula:
$$S= \int_{a}^{b} 2\pi*x*\sqrt{1+(\frac{dy}{dx})^2}$$
so here's my integral:

$$S= \int_{0}^{3\pi} 2\pi*x*\sqrt{1+(3x*cos(3x) + sin(3x))^2}$$

2.) The building will have a fabric exterior similar to a tent. The architect specifies that there must be a reinforcement in the wall for every 10 feet of wall (think arc length) and each end. How many reinforcement must be ordered?

Arc Length Formula = $$L = \int_{a}^{b} \sqrt{1+(\frac{dy}{dx})^2}$$

so y' = 3x*cos(3x) + sin(3x) , after that wouldn't i just plug it in the formula?

$$L = \int_{0}^{3\pi} \sqrt{1+(3x*cos(3x) + sin(3x))^2}$$

3.) The architect is hoping that people will see the building as waves in the ocean. She has specificed that the exposed exterior surface must be painted in sea-dappled blue paint. If the coverage for a gallon of sea-dapple blue paint. If the coverage for a gallon of sea-dapple blue paint is 100 square feet, how many gallons of paint are required? (then ends will open)

can i use shells to solve this? my integral
$$\int_{0}^{100} 2\pi*x*(x*sin(3x)+10)$$

can someone check if my setup for the integrals are correct?

 

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i posted this question really late (around 1AM), so i don't think anyone saw this question. so hopefully by moving it up, someone can help me. i really need help on these question, because i am stuck. if you can help on even one, that would help. thanks in advanve

 

[whozum]

1. Though confusing, I don't think you are looking for surface area. Its asking you to fill a volume in a certain amount of time. You know the base function, and you know the height. Integrate for volume.

2. Not really sure.

3. He wants to paint the walls. The area of the walls would be the surface area of building. Shells is a method for volume I believe.

 

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for #1.) when you mean integrate for volume, do you mean using shells?

and for #3.) do you mean that i should use both the surface area formula and shells to find the volume?

 

[whozum]

#1: Yes
#3: The volume is irrelevant in this problem. If you can find the surface area, you can find how much paint you need given the relationship 1 gallon = 100 ft^2

 

[AKG]

1. The pump must replace the entire volume of the building worth of air in 30 minutes, so find the volume of the building in cubic feet and divide by 30 to find the rate in ft³/min at which the pump has to work. To find the volume, you want to add the volumes of a bunch of semi-circular prisms of width dx and radius y(x), for x going from 0 to 3pi.

2. What is 1 foot of wall? I think you'd measure wall in area (ft²), not in length (ft), so the question isn't clear as written.

3. Find the surface area of the figure in square feet, divide by 100 to get the number of gallons required. To find the surface area, you need to find the surface area of the two ends, plus the surface area of the rest. To find it for the rest, you need to add the surface areas of a bunch of strips of with dx and length (2*pi*r)/2 (dividing by 2 because they're semi-circles) = pi*r = pi*y(x) (since y is the radius of the semi-circle) for the obvious range of x values.

You don't want to use shell method for anything. Shell method is a method for finding volumes, not areas, but you don't want to use shell method to find the volume here.

 

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for #1.) would the integral look like:
$$\frac{1}{30}*\int_{0}^{3\pi} (x*sin(3x)+10)^2$$

would i do 1/30 for 30mins or 1/2(for half an hour)?

and for #3.) surface area formula:
$$S= \int_{a}^{b} 2\pi*y*\sqrt{1+(\frac{dy}{dx})^2}$$


$$y' = 3xcos(3x)+sin(3x)$$

so...

$$S = 2\pi\int_{0}^{3\pi} {xsin(3x)+10}*{3xcos(3x)+sin(3x)}$$


did i do these correctly?