# integrate Area of a rt triangle w/ hypotenuse=r, horizontal=x, vertical=sqrt(r2-x2)

#### sinx

##### Junior Member
Area of a rt triangle, with the hypotenuse=r, horizontal=x, and vertical=sqrt(r2-x2)
I should get an equivalent to A=1/2base*ht

A=int sqrt (r2-x2)dx, limits for x=0,x
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin&
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
A=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}

questions;
1) I end up with r3, for an area??
2) what do i do with sin(rsin&)??

#### tkhunny

##### Moderator
Staff member
Area of a rt triangle, with the hypotenuse=r, horizontal=x, and vertical=sqrt(r2-x2)
I should get an equivalent to A=1/2base*ht

A=int sqrt (r2-x2)dx, limits for x=0,x
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin&
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
A=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}

questions;
1) I end up with r3, for an area??
2) what do i do with sin(rsin&)??
1) Your limits can't use the same variable as your argument. 'x' means two different things. Fix that.

2) Why are we doing sines and confusing ourselves? The fact that this is a simple geometry problem (Everything is straight!) and you already know the simple answer might suggest that something wandered off.

3) The equation of the hypotenuse, assuming the right angle is at the Origin, is $$\displaystyle \dfrac{x}{a} + \dfrac{y}{\sqrt{r^{2} - a^{2}}} = 1$$,where 'a' is the length of the base on the x-axis.

#### sinx

##### Junior Member
1) Your limits can't use the same variable as your argument. 'x' means two different things. Fix that.

2) Why are we doing sines and confusing ourselves? The fact that this is a simple geometry problem (Everything is straight!) and you already know the simple answer might suggest that something wandered off.

3) The equation of the hypotenuse, assuming the right angle is at the Origin, is $$\displaystyle \dfrac{x}{a} + \dfrac{y}{\sqrt{r^{2} - a^{2}}} = 1$$,where 'a' is the length of the base on the x-axis.
draw a rt triangle on x,y coordinate graph.
r is the hypotenuse, from 0 and at roughly 450
x is positive and to the right, from 0 to x.
y=sqrt (r2-x2​)

A=int dydx

#### Dr.Peterson

##### Elite Member
draw a rt triangle on x,y coordinate graph.
r is the hypotenuse, from 0 and at roughly 450
x is positive and to the right, from 0 to x.
y=sqrt (r2-x2​)

A=int dydx
No, y=sqrt (r2-x2​) is the equation of a semicircle, not a line!

As was pointed out, you seem to be using x to define the dimensions of the figure (the base), but also as the variable you are integrating. You can't do that.

These two facts together make your work nonsense; I wouldn't expect anything meaningful out of it.

Can you state the original problem you are trying to solve, exactly as given?

If you made it up, and just want to use calculus to find the area of a right triangle, then define things appropriately: base = b, height = h, so the hypotenuse is a line with slope h/b, and you can find the area by integrating y = (h/b)x from x=0 to x=b. The length of the hypotenuse won't come into the work at all.

#### sinx

##### Junior Member
No, y=sqrt (r2-x2​) is the equation of a semicircle, not a line!

As was pointed out, you seem to be using x to define the dimensions of the figure (the base), but also as the variable you are integrating. You can't do that.

These two facts together make your work nonsense; I wouldn't expect anything meaningful out of it.

Can you state the original problem you are trying to solve, exactly as given?

If you made it up, and just want to use calculus to find the area of a right triangle, then define things appropriately: base = b, height = h, so the hypotenuse is a line with slope h/b, and you can find the area by integrating y = (h/b)x from x=0 to x=b. The length of the hypotenuse won't come into the work at all.
y=sqrt (r2-x2​) defines y as function of r and x, pathagorus, where, r=hypotonuse
[it is also an equation of a circle with center at origin, where, r=radius]

my mistake was in treating r as a constant.
with r=f(x) this works.

#### Dr.Peterson

##### Elite Member
y=sqrt (r2-x2​) defines y as function of r and x, pathagorus, where, r=hypotonuse
[it is also an equation of a circle with center at origin, where, r=radius]

my mistake was in treating r as a constant.
with r=f(x) this works.
If you say so. It certainly can't be the quickest way to your goal, whatever that actually was.

Perhaps your ultimate mistake was in not clearly stating the problem and defining your variables.

#### sinx

##### Junior Member
If you say so. It certainly can't be the quickest way to your goal, whatever that actually was.

Perhaps your ultimate mistake was in not clearly stating the problem and defining your variables.
-don't be so quick to critisize.
-read the initial problem statement. It is easy enough to understand.
-you can use h for hypotenuse if r is confusing, and use b as the upper limit of x.

#### Dr.Peterson

##### Elite Member
-don't be so quick to critisize.
-read the initial problem statement. It is easy enough to understand.
-you can use h for hypotenuse if r is confusing, and use b as the upper limit of x.
Actually, I was asking you to show how you were able to solve it, and what result you got. It doesn't seem like an efficient way to do it, but perhaps it worked out better than I expect. And I was offering advice on how to avoid errors such as you know you made.

The problem statement was

Area of a rt triangle, with the hypotenuse=r, horizontal=x, and vertical=sqrt(r2-x2)
I should get an equivalent to A=1/2base*ht
The trouble with this is that it suggests that x and r are meant to be fixed numbers (base and hypotenuse of a triangle -- part of the statement of the problem), rather than variables to be used within the integration (which would not be part of the problem statement itself, but of the solution). You yourself stated that you initially treated r as a constant, but later realized it was not. Others pointed out that you used x as both the variable of integration, and a limit of integration, which is improper. These are the sorts of things that a full statement of the problem, together with a clear definition of any variables used, can prevent.

As I understand it, the problem statement would be, "Find the area of a right triangle with base ____ and height ____." I don't think you ever did identify variables for the base and height! Or, perhaps, you were aiming for a formula for the area of a triangle with base ____ and hypotenuse ____, which would be something like 1/2 b sqrt(h^2 - b^2). But, again, if r is a variable within the solution, what constant did you use for the entire hypotenuse? None of this is clear to me.

In particular, I'm curious why you would even want to mention the hypotenuse in a problem about area of a right triangle. (It's not the fact that you called it r, but that you didn't clarify its role in the problem, that was confusing.) Not that you can't, but it seems odd.

#### sinx

##### Junior Member
Actually, I was asking you to show how you were able to solve it, and what result you got. It doesn't seem like an efficient way to do it, but perhaps it worked out better than I expect. And I was offering advice on how to avoid errors such as you know you made.

The problem statement was

The trouble with this is that it suggests that x and r are meant to be fixed numbers (base and hypotenuse of a triangle -- part of the statement of the problem), rather than variables to be used within the integration (which would not be part of the problem statement itself, but of the solution). You yourself stated that you initially treated r as a constant, but later realized it was not. Others pointed out that you used x as both the variable of integration, and a limit of integration, which is improper. These are the sorts of things that a full statement of the problem, together with a clear definition of any variables used, can prevent.

As I understand it, the problem statement would be, "Find the area of a right triangle with base ____ and height ____." I don't think you ever did identify variables for the base and height! Or, perhaps, you were aiming for a formula for the area of a triangle with base ____ and hypotenuse ____, which would be something like 1/2 b sqrt(h^2 - b^2). But, again, if r is a variable within the solution, what constant did you use for the entire hypotenuse? None of this is clear to me.

In particular, I'm curious why you would even want to mention the hypotenuse in a problem about area of a right triangle. (It's not the fact that you called it r, but that you didn't clarify its role in the problem, that was confusing.) Not that you can't, but it seems odd.

I was trying to derive a simple equation for area of a rt triangle, using calculus and variables r and x, where r is hypotenuse, and x is horizontal.
(we secretly know it should be of the form A=1/2bh)

the integral to start with was A=int sqrt(r2-x2)dx, limits on x=0 to x (we can use b)
since r is dependent on x, we have to know r=f(x), assume r=2x.
A=int sqrt((2x)2-x2)dx=sqrt3intxdx, limits x=0 to b
A=1/2(sqrt3)b2
[or/ A=(1/2)base*ht for 600 triangle. i.e. r=2x]

we can also surmise other things we kinda know already, like
A=(b2/2) (tan&)

the reason why I wanted to do this is because it puzzled me.

#### Dr.Peterson

##### Elite Member
I was trying to derive a simple equation for area of a rt triangle, using calculus and variables r and x, where r is hypotenuse, and x is horizontal.
(we secretly know it should be of the form A=1/2bh)

the integral to start with was A=int sqrt(r2-x2)dx, limits on x=0 to x (we can use b)
since r is dependent on x, we have to know r=f(x), assume r=2x.
A=int sqrt((2x)2-x2)dx=sqrt3intxdx, limits x=0 to b
A=1/2(sqrt3)b2
[or/ A=(1/2)base*ht for 600 triangle. i.e. r=2x]

we can also surmise other things we kinda know already, like
A=(b2/2) (tan&)

the reason why I wanted to do this is because it puzzled me.
Okay, since you already know the area formula, I imagine that you are just experimenting, trying out alternative ways to find the area to see what happens and learn from it. Is that the context?

Evidently your idea was to suppose that you know, not the base and height (i.e. the two legs) as usual, but the hypotenuse and one leg. You wrote your integral, and has been already discussed, you did a couple things that made it confusing. You've ended up taking the base to be a known constant b, so that you are integrating with respect to x from 0 to b, and assuming that the angle at the origin is 60° (an arbitrary choice, to make the relation of r to x simple).

(If your goal was to obtain a general formula, you might write r = x/cos(θ), or just take the height h as known so that y = (h/b)x.)

So your integrand (which is just y, really) became sqrt(3)x, which could have been determined by seeing that for your triangle, h/b = sqrt(3).

If you'd made it general, such as r = x/cos(θ), then you would have found that the integrand is

y = sqrt((x/cos(θ))^2 - x^2) = sqrt(x^2(sec^2(θ) - 1)) = x tan(θ)

which would yield just the formula you mention. Of course, you could have gone more directly by starting out with y = x tan(θ).

Presumably you have, as planned, learned some things about pitfalls to look out for, and how to succeed. Experimentation is good!

#### sinx

##### Junior Member
Okay, since you already know the area formula, I imagine that you are just experimenting, trying out alternative ways to find the area to see what happens and learn from it. Is that the context?
yes,
and in this exercise there arose an integration puzzle.
how to integrate Z=int sqrt (r2-x2)dx? , where r is constant, limits for x=0, b

Z=int sqrt (r2-x2)dx, limits for x=0, b
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin& ??
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
Z=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}

then, with x/r=sin&
Z=(r2/4) [x+sinx]
units don't match, sinx is unitless.

#### Dr.Peterson

##### Elite Member
yes,
and in this exercise there arose an integration puzzle.
how to integrate Z=int sqrt (r2-x2)dx? , where r is constant, limits for x=0, b

Z=int sqrt (r2-x2)dx, limits for x=0, b
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&, limits 0, rsin& ??
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&, limits 0, rsin(1/2u)
Z=(r2/4) int (1+cosu) du;
integrate and evaluate @ limits 0, rsin(1/2u)
=(r2/4) {[rsin1/2u+sin(rsin1/2u)]-[0+0]}
=(r2/4) {[rsin&+sin(rsin&)]-[0+0]}

then, with x/r=sin&
Z=(r2/4) [x+sinx]
units don't match, sinx is unitless.
Keep in mind that this arose when you were wrongly taking r as a constant; so that your integral represents not a triangle, but part of a circle.

Also, you originally used x as the upper limit, and you are still doing that in your work. The upper limit, after substitution of "&", is not r sin& as you show (which would again be illegally using the variable with respect to which you are integrating), but & = arcsin(b/r). All along, your limits are wrong, which is what makes it impossible to untangle everything.

I would hold off on the limits, and just get the indefinite integral back in terms of x before doing the definite integral, to avoid mess.

#### sinx

##### Junior Member
I would hold off on the limits, and just get the indefinite integral back in terms of x before doing the definite integral, to avoid mess.
Z=int sqrt (r2-x2)dx, limits for x=0, b
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&,
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&,
=(r2/4) int (1+cosu) du
Z=(r2/4) (u+sinu) l [indefinite integral]

_______________________________

u=2sin-1(x/r),
limits for x=0, b
upper limit=
2sin-1(b/r), lower limit=0
z=(r2/4) (2sin-1b/r+sin(2sin-1b/r))-(0+0)
Z=(r2/4) (2sin-1b/r+sin(2sin-1b/r))

how do I simplify sin(
2sin-1b/r)? ..................use trig identity sin2A=2sinAcosA???
Z=(r2/4) (2sin-1b/r+2(b/r)sqrt(1-(b/r)2))???

Z=(r2/2) (sin-1b/r+(b/r)sqrt(1-(b/r)2)???

this does not look right, I have two terms with different units.
the first is ok, because it is an angle in radians, but the second does not have units.
they should both be in radians, or maybe b/r could be thought of as radians, like pi= circ/dia ??

Last edited:

#### Dr.Peterson

##### Elite Member
Z=int sqrt (r2-x2)dx, limits for x=0, b
=r int sqrt (1-(x/r)2)dx;
sub x/r=sin&, dx=rcos&d&,
=r2 int cos2& d&;
sub cos2&+1=2cos2&
=(r2/2) int (1+cos2&) d&;
sub u=2&, , du=2d&,
=(r2/4) int (1+cosu) du
Z=(r2/4) (u+sinu) l [indefinite integral]

_______________________________

u=2sin-1(x/r),
limits for x=0, b
upper limit=
2sin-1(b/r), lower limit=0
z=(r2/4) (2sin-1b/r+sin(2sin-1b/r))-(0+0)
Z=(r2/4) (2sin-1b/r+sin(2sin-1b/r))

how do I simplify sin(
2sin-1b/r)? ..................use trig identity sin2A=2sinAcosA???
Z=(r2/4) (2sin-1b/r+2(b/r)sqrt(1-(b/r)2))???

Z=(r2/2) (sin-1b/r+(b/r)sqrt(1-(b/r)2)???

this does not look right, I have two terms with different units.
the first is ok, because it is an angle in radians, but the second does not have units.
they should both be in radians, or maybe b/r could be thought of as radians, like pi= circ/dia ??

Radians should be thought of as dimensionless, since they are defined as a length over a length. So there is no conflict. (The term "radian" is retained mostly as a reminder of where a number came from.)

Good work. Everything looks right (except for some omitted parentheses).

#### sinx

##### Junior Member
Radians should be thought of as dimensionless, since they are defined as a length over a length. So there is no conflict. (The term "radian" is retained mostly as a reminder of where a number came from.)

Good work. Everything looks right (except for some omitted parentheses).
thank you