# Volume Problems

1. Nov 11, 2008

### nns91

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
Integrals.

3. The attempt at a solution

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

I am totally lost for others. Really URGENT

2. Nov 11, 2008

### 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.

3. Nov 11, 2008

### 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 ??

4. Nov 11, 2008

### 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?

5. Nov 11, 2008

### nns91

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

6. Nov 11, 2008

### Dick

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

7. Nov 11, 2008

### nns91

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

8. Nov 11, 2008

### nns91

Stupid me it is pi/2.

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

9. Nov 11, 2008

### abelanger

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!

10. Nov 11, 2008

### nns91

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 ?

11. Nov 11, 2008

### 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 dont know if what you have is right though.

12. Nov 11, 2008

### 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 ?

13. Nov 11, 2008

### abelanger

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)) ??

14. Nov 11, 2008

### Dick

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?

15. Nov 11, 2008

### 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 ?

16. Nov 11, 2008

### 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.

17. Nov 11, 2008

### 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 ?

18. Nov 11, 2008

### 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.

19. Nov 11, 2008

### 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 ?

20. Nov 11, 2008

### 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).