Help me find the upper limit of integration

[ArtVandelay]

Homework Statement

The volume of a tank, a half-cylinder with the flat side UP, when it is FULL, is given by the following:$$16 \int_0^2 \sqrt{4-y^2} dy$$ where 0 is the top of the tank and 2 is the bottom. We also know that when evaluated from 0 to 1 (depth of gas = 1), it equals the following: $$16 \int_1^2 \sqrt{4-y^2} dy\ = \ 8/3 (4 \pi-3 \sqrt{3})\ \approx\ 19.6539$$ Now imagine that that same volume of gas (19.6539) remains in the tank, but the tank is flipped upside down (so now the flat side is on the bottom). Find the depth of the gas.

Homework Equations

Hints from instructor:
* Do not try to integrate $\sqrt{4-y^2}dy$
* Use a calculator

The Attempt at a Solution

I'm pretty sure I just need to solve for k in the following: $$16 \int_0^k \sqrt{4-y^2} dy\ = \ 19.6539$$ But I don't know how to do that.

 

[sgnmath]

This is ?. It is much nicer if you (or WE) are allowed to integrate.
But TI89:
nsolve(nInt(16 sqrt(4-y^2),y,0,x)=19.6539,x)
where x is your k.

nSolve is under F2, nInt is under F3,
nSolve(nInt(16 \sqrt(4-y^2), y,0,x) = 19.6539, x).

You might want to do it with without n, but it takes longer (if it doesn't crash).
IF you want to know how to do this on TI83 or TI84, then be nice!
SKYLINE

 

[Ray Vickson]

Homework Statement

The volume of a tank, a half-cylinder with the flat side UP, when it is FULL, is given by the following:$$16 \int_0^2 \sqrt{4-y^2} dy$$ where 0 is the top of the tank and 2 is the bottom. We also know that when evaluated from 0 to 1 (depth of gas = 1), it equals the following: $$16 \int_1^2 \sqrt{4-y^2} dy\ = \ 8/3 (4 \pi-3 \sqrt{3})\ \approx\ 19.6539$$ Now imagine that that same volume of gas (19.6539) remains in the tank, but the tank is flipped upside down (so now the flat side is on the bottom). Find the depth of the gas.

Homework Equations

Hints from instructor:
* Do not try to integrate $\sqrt{4-y^2}dy$
* Use a calculator

The Attempt at a Solution

I'm pretty sure I just need to solve for k in the following: $$16 \int_0^k \sqrt{4-y^2} dy\ = \ 19.6539$$ But I don't know how to do that.

(1) Work out a formula for the integral
$$F(k) = 16 \int_0^k \sqrt{4-y^2} dy $$
in terms of k (just using standard integration methods, or tables, or whatever).
(2) Solve the equation F(k) = 19.6539 numerically, using one of the many methods available, such as Newton's method. Plotting a graph to get a rough idea of the root is a good way to start.

Alternatively, use a computer solving package, such as the EXCEL Solver.

RGV

 

[sgnmath]

That doesn't help. They are not supposed to integrate!

 

[Ray Vickson]

That doesn't help. They are not supposed to integrate!

OK, I missed that.

RGV

 

[ArtVandelay]

Thanks guys. I was able to do it on a TI83 by graphing y = 19.65 and y = [integral], then finding the intersection. Then, five minutes later, I successfully repeated that operation on my midterm :)

 

[sgnmath]

OK. Good.
Since you figured that out. Here is the math. Find k so that
$$\int_o^k 16 \sqrt(4-x^2) dx = \int_1^2 16 \sqrt(4 - x^2)dx = 19.6539$$.
The first integral is not a constant, but the second one is.
To be precise (and to see technology can be slo), with TI83 or 84: in Y1, type the first integral; in Y2, type the second integral:
Y1=fnInt(16 √(4-x^2), X, 0, X)
Y2=fnInt(16 √(4x-x^2), x, 1, 2)
To save 5 minutes, you can just type Y2 = 19.6539.
Then use the "intersect" feature: 2nd TRACE 5.

 

[Ray Vickson]

Thanks guys. I was able to do it on a TI83 by graphing y = 19.65 and y = [integral], then finding the intersection. Then, five minutes later, I successfully repeated that operation on my midterm :)

So, you let your II83 do the integral? I thought you said you were not supposed to use integration!

RGV

 

[sgnmath]

from the original post:
"
* Do not try to integrate \sqrt(4-x^2).
* Use a calculator
"

 

[Ray Vickson]

from the original post:
"
* Do not try to integrate \sqrt(4-x^2).
* Use a calculator
"

OK, but that is a ridiculous way to teach mathematics and an insane requirement in an assignment.

RGV

 

[sgnmath]

hence my comment in my first post: This is ?, where initially ? was Dumb but I changed it.

 

[ArtVandelay]

OK, but that is a ridiculous way to teach mathematics and an insane requirement in an assignment.

RGV

To the instructor's credit, these past few weeks were really heavily focused on the conceptual idea of using integration to find volumes of various shapes / functions. He really wanted to keep the number-crunching part down to a minimum. 95% of the problems we did involved definite integrals that we could evaluate by hand if we pleased. This problem was the exception (i.e. we have not yet learned how to find a bound for an integral when we know what it equals).

This is 7 weeks into Calc II, btw.

 

[sgnmath]

I totally agree with the last post. I thought there was a trick or an elegant way to do it (while avoiding doing integration by hand), but there was none. It is dumb only when I realized that getting the answers involves knowing which command/syntax to use in TI calculator.