Volume Problems: Solving for Area and Volume with Integrals

[nns91]

Homework Statement

1. The base of a solid S is the shaded region in the xy-plane enclosed by the x-axis, the y-axis and the graph of y=1-sinx. For each x, the cross section of S perpendicular to the x-axis at the point (x,0) is an isosceles right triangle whose hypotenuse lies in the xy-plane.

(a) Find the area of the triangle as a function of x.

(b) Find the volume of S.


2. An oil storage has the shape obtained by revolving the curve y=9*x^4/625 from x=0 to x=5 about the y-axis, where x and y are measured in feet. Oil flows into the tank at the constant rate of 8 cubic feet/min.

(a) find the volume of the tank. Indicate units of measure.

(b) To the nearest minute, how long would it take to fill the tank if the tank was empty initially ?

(c) Let h be the depth ,in feet, of oil in the tank. How fast is the depth of the oil in the tank increasing when h=4 ? Indicated units of measure.

Homework Equations

Integrals.

The Attempt at a Solution

For number one, I found that the length of the triangle is 1-sinx. so is the area just (1-sinx)^2/2 ??

I am totally lost for others. Really URGENT

 

[Dick]

If this is URGENT, you'd better get busy. You've already got a problem with 1a). 1-sin(x) is the length of the hypotenuse of the right triangle. Draw one. What's the area? The sooner you present an attempt at a solution for the other parts, the sooner you will get help.

 

[nns91]

so side is 1-sin(x)/ sqroot(2) ? Thus, the area will be (1-sin(x)^2/4 ?

For part b, will it be the integral from 0 to 1.571 ??

 

[Dick]

Yes, that's the area. You mean the integral of the area, right? Sure. But I think they will want you to give an exact value of the limit instead of the approximation 1.571. Did you get that from a graphing calculator or something?

 

[nns91]

Yeah, I did. Which can be a more accurate way to find out the x interception ?

 

[Dick]

The x intercept is where y=0. Solve y=1-sin(x)=0. Look at a graph of sin.

 

[nns91]

I got a really weird thing from my TI-89. Should I use the approximation instead then ?

 

[nns91]

Stupid me it is pi/2.

I am really lost at the 2nd problem. Can you help me ?

 

[abelanger]

Something that might help you:
Area from a-->b = integral from a-->b of f(x) dx
Volume from a-->b = (pi)(integral from a-->b of f(x)squared dx)
So for #1 find the integral of: 1-sinx dx (from 0 to pi/2)
See what you get!

 

[nns91]

I got about 0.089.

Ok. I got number 1.

But number 2 I am kinda lost. Should I use shell or dish method here ? If I use dish, I have limits of integration from 0 to 5 but what can be the integrand ?

If I use shell. I will have x= (y*625/9)^(1/4). Thus, my formula will be something like

pi* integral ( x^2, evaluation from 0 to 9) ?? Is it true ? Which way should I follow ?

 

[abelanger]

Ok first make sure if you use your calculator that it is in radians... because that is not what I got for the area of #1. By dish method do you mean disk lol? Anyways, since you revolve around the y axis, you have to put your function in terms of x =, now i don't know if what you have is right though.

 

[nns91]

What did you get ? That's what I got in radian mode. Did you square (1-sin(x)) ? and then divide the integral by 4 ?

What do you mean ? I meant disk, sorry.

I did put function in term of x like above. is that right ?

 

[abelanger]

Sorry got back from lunch,

What did you get ? That's what I got in radian mode. Did you square (1-sin(x)) ? and then divide the integral by 4 ?

I did put function in term of x like above. is that right ?

Why would i square 1-sin(x) and then divide the integral by 4?? do you know how to get the integral of : f(x) = 1-sin(x) ??
Well is the equation: (9(x^4))/625 OR (9)(x^(4/625)) ??

 

[Dick]

Sorry got back from lunch,

Why would i square 1-sin(x) and then divide the integral by 4?? do you know how to get the integral of : f(x) = 1-sin(x) ??
Well is the equation: (9(x^4))/625 OR (9)(x^(4/625)) ??

You would do those things because that how you get the cross sectional area to integrate. nns91 is doing it correctly, the answer is about 0.089. I don't think you read the problem completely. But, again, nns91, I think you want to express the answer exactly instead of as an approximation. Can't you do the integral without doing it numerically?

 

[nns91]

I could but I think my teacher prefer approximation. Thanks for your comment.

The problem is (9*x^4)/625.

So how can I do the 2nd problem ?

 

[Dick]

Do the next one like you did the first one. Sure, use disks. Given a value of y, what's the radius of the disk? What's the area of the disk? Now integrate area dy to get the volume.

 

[nns91]

Radius will be x=((625*y)/9)^1/4 so area will be pi*x^2 ? Is that true ?

I am confused about part c too.

So v= pi*r^2 * h right ? What's next ? I too the derivative but it led to dr/dt which I don't know ?

 

[Dick]

Right. Area is pi*x^2=pi*((y*625)/9)^(1/4))^2. But you don't just multiply area times height, the area isn't constant. You have to INTEGRATE area*dy. Because you know part c) is coming don't just find the volume for h=5. Find the volume for a general value of h.

 

[nns91]

so how can I find the limit of integral is that 0 to 9 ?

v=pi*r^2*h right ? so I just take the derivative. and gt dv/dt = 2pi*r*h*dr/dt + 2pi*r*dh/dt.

I have pretty much everything except dr/dt and r. How can I find those variables ?

 

[Dick]

Maybe I didn't emphasize that V=pi*r^2*h is WRONG. You have to INTEGRATE to find the volume. Go back to part a).

 

[nns91]

Oh ok. So how can I process it ??

 

[Dick]

Integrate area as a function of y from y=0 to y=h.

 

[nns91]

and then take the derivative ??

 

[nns91]

so I got v= (50*h^(3/2*pi)/9.

v'(t)= 25pi/3* h^1/2* h'(t).

Then I solve for h'(t) and I got a really small number like 0.153. Does it make sense ??

 

[Dick]

You've got some parentheses out of place, but I get what you mean. And, yes, I seem to get a numerical answer for h'(t) of 0.153. You might want to notice that the question asks you to put correct dimensions on things. How do you know your number is small if you don't put dimensions on it?

 

[nns91]

it is ft/min ?

 

[Dick]

it is ft/min ?

Sure it is. You really didn't have to ask that, right? Now figure out if your number is reasonable. The tank contains a couple hundred cubic feet of oil and is 5 feet tall. Given that does 0.15ft/min seem like it's in the right ballpark for 8 cubic feet per minute input?

 

[nns91]

I guess ??

 

[Dick]

You mean "It think it is.". You don't have to put '??' after every statement. You solved the problem pretty easily once you got a grip on it.

 

[nns91]

I think it is. Do you ?

 

[Dick]

Sure. Estimate with rounded numbers. 200ft^3/(8*ft^3/min)=25 minutes. 5 ft/25 min=0.2ft/min roughly on average. That makes 0.15ft/min sound pretty reasonable. Not 'too small'.

 

[nns91]

Thanks for your help

 

[Dick]

Thanks for your help

Very welcome!