Definite Double Integrals 2 variables TI-89 Titanium

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1. Apr 18, 2014

Digitalism

Hello I am trying to solve this integral 25-9x^2-25y^2/9 dydx integrating from 0 to sqrt(9-9x^2/25) and the limits of the second integration are 0 to 5. I can find tutorials on how to find the definite double integral of a single variable, but not for two variables. Any clues?

edit: so far I have tried nint(nint(25-9x^2-25y^2/9,y,0,sqrt(9-9x^2/25)),x,0,5) to which I get too few arguments and int(int(25-9x^2-25y^2/9,y,0,sqrt(9-9x^2/25)),x,0,5) which also results in "too few arguments." Thanks in advance!

Last edited: Apr 18, 2014
2. Apr 18, 2014

Staff: Mentor

I don't see any reason yet for the "too few arguments" error you're getting, but I'm still looking.

I see a couple of other things that probably are errors, though. I'm guessing that your integrand is supposed to be $\frac{25 - 9x^2 - 25y^2}{9}$. However, what you wrote is $25 - 9x^2 - \frac{9y^2}{9}$. You would need to write your integrand as (25 - 9x^2 - 25y^2)/9 if the first expression above is what you intended.

You have the same problem with one of your limits of integration. Inside the radical, what you wrote is 9 - (9x^2/25). I believe you meant (9 - 9x^2)/25.

3. Apr 18, 2014

Staff: Mentor

nint won't work for you, as it gives an approximation. And int(<something>) doesn't do integration. From what I can tell from the TI documentation, you need to enter the integral symbol (∫).

On page 902 of the TI-89 Guidebook (Appendix A Functions and Instructions), they show an example of a double integral.

What you enter into the calculator looks like this:
∫(∫(ln(x + y), y, 0, x), x, 0, a) <ENTER>
After pressing the ENTER key, the display should look like this:
$$\int_0^a \int_0^x ln(x +y)dy~dx$$

4. Apr 18, 2014

Digitalism

sorry I wasn't doing int on the calulator I was using the integral function but I didn't know how to represent that here on the forums.

Also I did intend 25-9x^2-25y^2/9

Here is the problem:

Find the volume of the solid by using an iterated integration. Solid in the first octant, bounded by the surface
9z = 225 -9x^2-25y^2

//end question

My method:

setting z=0 and solving for y

25y^2=225-9x^2
y= +/-sqrt(9-9x^2/25)

setting y = 0
x=+/-5

because the volume is bounded by the first octant the limits of integration should be 0<=x<=5 and 0<=y=sqrt(9-9x^2/25)

yes?

so z = 25-x^2-25y^2/9

or did I do that incorrectly?

edit:

I tried this with the function integral(integral(xy),y,0,x),x,0,5) just to try a different function and I am getting the same error

note the integral function I am using on the TI-89 titanium is the 2nd option under F3 just to make sure we're using the same integral function. I saw that in the manual as well and I cannot explain the error.

Last edited: Apr 18, 2014
5. Apr 19, 2014

Staff: Mentor

Intended or not, it's incorrect.
So $z = \frac{225 -9x^2-25y^2}{9}$.
What you wrote, and entered into the calculator, was
$z = 225 -9x^2-\frac{25y^2}{9}$.
The difference should be obvious.
When you have to write a fraction on a single line, and the numerator or denominator (or both) has two or more terms, you HAVE to put parentheses around it (them), like so:
(225 - 9x^2 - 25y^2)/9
Since the denominator consists of only one term, parentheses aren't needed for it.
Incorrect for two reasons.
1. The quantity inside the radical has the same problem as shown above.
2. Since the "cap" over the region is in the first octant, all coordinates are ≥ 0.

Written correctly, you should have y = sqrt((9 - 9x^2)/25). Notice that there are two pairs of parentheses - one pair for the sqrt function, and another pair inside for the numerator of the fraction.

Omitting the +/-, this is what you wrote:
$$y = \sqrt{9 - \frac{9x^2}{25}}$$
If this is exactly what you entered into the calculator, the main problem I see is that your parentheses aren't matched correctly. You have two left parens - ( - and three right parens - ).
The one in red should go away: integral(integral(xy),y,0,x),x,0,5)

6. Apr 19, 2014

Digitalism

Sorry when I first posted the 9 in front of the x^2 term was a typo.

Wait, according to what you wrote above shouldn't it be

$$y = \sqrt{\frac{225 -9x^2}{25}}$$

Ok thanks! Sorry when I get in a rush I make sloppy mistakes like that.

7. Apr 19, 2014

Staff: Mentor

Yes. That was a mistake on my part. And this simplifies to what you had earlier. I didn't notice that you had already divided the 225 by 25.

8. Apr 19, 2014

Digitalism

It worked! Thanks!

9. Apr 19, 2014

Staff: Mentor

Glad to hear it! I don't have a TI calculator. I have an HP 48G that I haven't looked at for 15+ years. What I did in this thread was to download the TI89 Titanium Guidebook from the website. They have examples of the various operations the calculator can do. If you don't already have this documentation, I would recommend downloading it (it's free).