Trigonometric Substitution- Area help

[Slimsta]

Homework Statement

Let R be the smaller of the two regions enclosed by the elipse 144 x²+64 y²=9216 and the line $$x=(8 \sqrt{2})/2$$.
Find the area of the region R.

Homework Equations

The Attempt at a Solution

my textbook doesn't have anything like this.. i have no idea where to start even.
please help!

 

[Dick]

Sketch the curve and the line. This is really just a plain 'area under the curve problem'. There's nothing exotic about it.

 

[Slimsta]

her is the image of the graph:
http://img521.imageshack.us/img521/5048/ellipse.png

so x_R = 4*$$\sqrt{2}$$
x_L = $$\sqrt{(576-4y^2)/9}$$ (i reduced down all the numbers of the elipse by 16)

and now just find A(x) = $$\int_{b'}^b (XR - XL)dx$$

?

 

[Dick]

Well, ok. What you are describing is how you would do it by integrating dy, not dx. But how can you 'reduce numbers by 16'?? Don't do that. x=sqrt(9216-64*y^2). Now you just need to find b and b' and do the integration. You might find it a little easier to integrate dx instead. Solve for y in terms of x and integrate from x=4*sqrt(2) to the end of the ellipse.

 

[Slimsta]

Well, ok. What you are describing is how you would do it by integrating dy, not dx. But how can you 'reduce numbers by 16'?? Don't do that. x=sqrt(9216-64*y^2). Now you just need to find b and b' and do the integration. You might find it a little easier to integrate dx instead. Solve for y in terms of x and integrate from x=4*sqrt(2) to the end of the ellipse.

well in other words, i did everything right except the 'reducing by 16' part.. right?
all i got to do now is plug my x's into the
A= $$\int_{b'}^b (XR - XL)dx$$

find the antiderivative and I am good to go... correct me if I am wrong please!

 

[Dick]

Right, except doing it the way you've outlined means you are integrating dy. You need to find y bounds.

 

[Slimsta]

Right, except doing it the way you've outlined means you are integrating dy. You need to find y bounds.

ok I am lost now.
there is this equation for elipse (x^2 / a^2) + (y^2 / b^2) = 1
so i get (x^2 / 64) + (y^2 / 144) = 1
then i solve for y and i get
y = +- sqrt(144- (9x^2)/4 )
so the y+ is the upper bound and y- is the lower bound.

apparently I am supposed to use the substitution 'asint' well because its Trigonometric Substitution lol

what confused me is that you said I am right but then the students from my class say other things on the online assignment website.

A = $$\int_{-\sqrt(144- (9x^2)/4 )}^{\sqrt(144- (9x^2)/4 )} (\sqrt(9216-64*y^2) - 4*\sqrt{2})dy$$

or am i supposed to do y_T - y_B? because from the graph it looks like it should be the X's not Y's.. if you know what i mean by that

 

[Dick]

You can do it either dx or dy. It will work either way. Your problems now are i) you are STILL writing dx instead of dy and ii) your upper and lower limits should be numbers. They are the y coordinates of the points where the line and the ellipse intersect.

 

[Slimsta]

oh right i forgot to change it to dy..
i was thinking doing this:
2*A = $$\int_{0}^{\sqrt(144- (9x^2)/4 )} (\sqrt(9216-64*y^2) - 5.65685)dy$$

how do i get a # for the upper bound? i mean what x-value do i plug in there?
im happy that I am on the right truck at least lol

 

[Dick]

To find the bounds you intersect x=4*sqrt(2) with the ellipse. Just substitute x=4*sqrt(2) into the ellipse equation and find the corresponding y.

 

[Slimsta]

oh okay.. so i get
2A = $$\int_{0}^{8.48528} (\sqrt(9216-64y^2)dy - 48 + C$$

y = 96sint
dy = 96cost dt

==>
2A = $$\int_{0}^{8.48528} [\sqrt(9216-64(96sint)^2]*96cost dt - 48 + C$$

something is wrong.. i feel like I am missing a step or something, can you tell me what am i missing?

 

[Dick]

Something is wrong. Why do you think 96*sin(t) is a good substitution? And can you show more of your steps? I had a hard time figuring out where the 48 came from. Especially since you are changing limits to decimal. And there's no call for a '+C'. These are definite integrals.

 

[Slimsta]

Something is wrong. Why do you think 96*sin(t) is a good substitution? And can you show more of your steps? I had a hard time figuring out where the 48 came from. Especially since you are changing limits to decimal. And there's no call for a '+C'. These are definite integrals.

sorry about that.. i just got excited when i got the answer for that part so i just put the number in.
$$\sqrt(9216-64y^2)$$ = $$\sqrt(96^2-(8y)^2)$$

and as that substitution rule states: $$\sqrt(a^2-x^2)$$
=> x= asint
dx = acostdt
so $$\sqrt(a^2-(asint)^2)$$

so..
y= 96sint
dy = 96costdt
$$\sqrt(96^2-(8*96sint)^2)$$

which leads to
$$\int_{0}^{8.48528} [\sqrt(9216-(8*96sint)^2]*96cost dt - 48$$

 

[Dick]

The 'x' in your subsitution rule is 8y. You want 8y=96*sin(t).

 

[Slimsta]

ok so i just found a mistake in my X_R calculation.
it's actually
$$\sqrt((9216-64y^2)/144)$$ = $$\sqrt(64-(4y^2)/9)$$ =

$$\sqrt(8^2-(2y/3)^2)$$

cuz i forgot to divide the whole thing by 144.

so now, 2y/3 = 8sint
y=12sint
dy = 12costdt

A= $$\int_{-8.48528}^{8.48528} ([\sqrt(64-(8sint)^2 ] - 4\sqrt(2) ) dy$$

which leads to
$$\int_{-8.48528}^{8.48528} (8cost - 4\sqrt(2) ) dy$$
=$$\int_{-8.48528}^{8.48528} (8cost - 4\sqrt(2) ) 12costdt$$
=12 $$\int_{-8.48528}^{8.48528} (8cos^2t - 4\sqrt(2) cost)dt$$
=12*4 $$\int_{-8.48528}^{8.48528} (2cos^2t - \sqrt(2) cost)dt$$
=48*( $$2 *\int_{-y}^{y} (cos^2t)dt - \sqrt(2)*\int_{-y}^{y} ( cost)dt$$)

=48*( $$2 *\int_{-y}^{y} (0.5(1-cos2t))dt - \sqrt(2)*sint$$)

=48*( $$\int_{-y}^{y} (1-cos2t)dt - \sqrt(2)*sint$$)

is anything wrong with it?
i checked it over many times already..
my question is now what do i do once i find the antiderivatives.. i mean, change sin/cos with whatever i substituted it to?

 

[Dick]

You are really losing me here. How did the trig functions get mixed up the the 4*sqrt(2) part? I thought you already decided that was 48? Why don't you back up and make a fresh start and present all of your steps in order. I really can't follow that. I just can't. You are doing some stuff right, but the organization is just so messed up.

 

[Slimsta]

x_R = $$\sqrt((9216-64y^2)/144)$$
x_L = 4 $$\sqrt{2}$$ = 5.65685

i got the bounds a = -8.485, b = 8.485

A=$$\int_{a}^{b} (xr-xl)dy$$

A=$$\int_{a}^{b} ((\sqrt((9216-64y^2)/144))-5.65685)dy$$

A=$$\int_{a}^{b} ((\sqrt((64-(4/9)y^2)))-5.65685)dy$$

A=$$\int_{a}^{b} ((\sqrt((8^2-(2y/3)^2)))-5.65685)dy$$

A=$$\int_{a}^{b} \sqrt(8^2-(2y/3)^2)dy - \int_{a}^{b}5.65685dy$$

(2y/3) = 8sint
y=12sint
dy=12costdt

A=$$\int_{a}^{b} \sqrt(64-(8sint)^2)dy - 5.65685y|_{a}^{b}$$

A=$$\int_{a}^{b} \sqrt(64-64sin^2t dy - 5.65685y|_{a}^{b}$$

A=$$\int_{a}^{b} \sqrt(64(1-sin^2t)) dy - 5.65685y|_{a}^{b}$$

A=$$\int_{a}^{b} \sqrt(64(cos^2t)) dy - 5.65685y|_{a}^{b}$$

A=$$\int_{a}^{b} 8costdy - 5.65685y|_{a}^{b}$$

A=$$8sint|_{a}^{b} - 5.65685y|_{a}^{b}$$

recall that (2y/3) = 8sint
so
A=$$2y/3 |_{a}^{b} - 5.65685y|_{a}^{b}$$

and i get A=84.6862 but its wrong!

if i do 0.5A=$$2y/3 |_{0}^{8.485} - 5.65685y|_{0}^{8.485}$$ i still get 84.6862

 

[Dick]

Now you dropped the cos(t) from the dy=12costdt, and you've gone back to changing xR from sqrt(9216-64y^2) to sqrt((9216-64y^2)/144). They AREN'T the same thing. You are changing things you corrected back into the old incorrect things. If you are as tired as I am right now I think you should get some rest and try to sort things out in the morning.

 

[Slimsta]

i don't have til the morning.. its due 8am :'(
how did you get sqrt(9216-64y^2) in the first place?

$$\int_{a}^{b} \sqrt(8^2-(2y/3)^2)*12costdt- (5.65685)*\int_{a}^{b} 12costdt$$

$$12*\int_{a}^{b} \sqrt(64 - 4y^2/9)costdt - (67.88)sint |_a^b$$

im really sorry for the headache i give you. our teacher just expects us to know all this without teaching us so what can i do..

 

[Dick]

Oops, you got me. Sure xR=sqrt(64-(2y/3)^2). The problem is I keep flipping back to your original post and the 144 is on the first line and I keep forgetting it's there. Sorry. But I'm still having a hard time tracking here. I am integrating sqrt(64-(2y/3)^2)-4*sqrt(2))*dy from -6*sqrt(2) to +6*sqrt(2) and I'm not getting 84.6862. That's about all I've got in me right now.

 

[Slimsta]

Oops, you got me. Sure xR=sqrt(64-(2y/3)^2). The problem is I keep flipping back to your original post and the 144 is on the first line and I keep forgetting it's there. Sorry. But I'm still having a hard time tracking here. I am integrating sqrt(64-(2y/3)^2)-4*sqrt(2))*dy from -6*sqrt(2) to +6*sqrt(2) and I'm not getting 84.6862. That's about all I've got in me right now.

well see.. then i did everything correct, can you tell me what you get? i must of punched it wrong in the calculator or something. wow this question takes more time than i thought.

im stuck over here:
48* ( $$\int_a^b (1-cos2t)dt - \sqrt(2)*sint|_a^b$$)

48* $$(t - (sin2t)/2 - \sqrt(2)*sint )|_a^b$$

t=sin^{-1}(y/12) and sint = y/12 so

48* $$(sin^{-1}(y/12) - (sin2t)/2 - \sqrt(2)*(y/12) )|_a^b$$

what do i do with sin2t ?

 

[Dick]

You know the y limit is 6*sqrt(2). What's the t limit? Or you can just write sin(2t)=2*sin(t)*cos(t) and express cos(t) and sin(t) in terms of y.