Thread: Area under curve, using parameters.

1. Area under curve, using parameters.

The diagram shows the curve defined by parametric equations $\displaystyle x = t^{2}$ , $\displaystyle y = t^{3}$

for values of t between -3 and 3 find the shaded area.

I have drawn the diagram shows in my book and this my set up,

$\displaystyle \int^3_{-3} t^{3} \\\ 2t \\\ dt$

However when I compute this, I do not get the right answer. Is this the correct method?

thank you.

http://www.mathhelpforum.com/math-he...8&d=1275589381

2. So long as the curve is traversed (I always hated that word; travelled sounds better) once, then you can use direct substitution as you have done. What is the answer the book gets?

3. Originally Posted by ANDS!
So long as the curve is traversed (I always hated that word; travelled sounds better) once, then you can use direct substitution as you have done. What is the answer the book gets?
I dont know what you mean by 'transversde' in relation to this graph?

And the book says the correct answer is 110, and I keep getting 194.4

4. Traversed means the line is traced out only once from -3 to 3. Are you sure you have your limits of integration correct, as well as the correct equations for x and y? I get the answer as you do using the information you have provided.

5. Originally Posted by ANDS!
Traversed means the line is traced out only once from -3 to 3. Are you sure you have your limits of integration correct, as well as the correct equations for x and y? I get the answer as you do using the information you have provided.

The limits if integration are given to us aren't they? We dont have to work it out. the question says values of between -3 and 3.

6. I get the same answer you are getting. If they say -3 to 3, then those are the limits of integration. There must be some bit of information missing - do you have the original problem as written (without omissions?).

7. Originally Posted by ANDS!
I get the same answer you are getting. If they say -3 to 3, then those are the limits of integration. There must be some bit of information missing - do you have the original problem as written (without omissions?).

I have posted the original question word for word. I am pretty sure the book is not wrong, its always right!

8. Originally Posted by Tweety
The diagram shows the curve defined by parametric equations $\displaystyle x = t^{2}$ , $\displaystyle y = t^{3}$

for values of t between -3 and 3 find the shaded area.

I have drawn the diagram shows in my book and this my set up,

$\displaystyle \int^3_{-3} t^{3} \\\ 2t \\\ dt$

However when I compute this, I do not get the right answer. Is this the correct method?

thank you.

http://www.mathhelpforum.com/math-he...8&d=1275589381
note the attached parametric graph from t = -3 to t = 3

your sketch looks like it starts at t = -2

$\displaystyle x = t^2$

$\displaystyle y = t^3$

$\displaystyle y = x^{\frac{3}{2}}$

$\displaystyle A = 2\int_0^9 x^{\frac{3}{2}} \, dx$

9. Originally Posted by skeeter
note the attached parametric graph from t = -3 to t = 3

your sketch looks like it starts at t = -2

$\displaystyle x = t^2$

$\displaystyle y = t^3$

$\displaystyle y = x^{\frac{3}{2}}$

$\displaystyle A = 2\int_0^9 x^{\frac{3}{2}} \, dx$

But the sketch in my book says t = -2 and t = 3, is this a misprint than? And they were suppose to put -3 instead of -2?

10. Originally Posted by Tweety
But the sketch in my book says t = -2 and t = 3, is this a misprint than? And they were suppose to put -3 instead of -2?
I don't know, I can't see your book.

I would go with the picture. Re-evaluate from t = -2 to t = 3.

11. I probably should have looked at that picture. The limits of integration are -2 to 3.