Evaluate the integral by interpreting it in terms of areas.

Hemolymph
Messages
30
Reaction score
0

Homework Statement



from [-2,2] rad(4-x^2)

Homework Equations





The Attempt at a Solution


I know its a circle and i get the equation to be y^2+x^2=4

and I believe it has to be divided into a circle and rectangle
so the area of the rectangle i got to be 2
the circle i got to be 1/2(since its a half circle) times ∏(2)^2
the answer i got is
2+2pi (which is wrong) don't know where I went wrong tho.
 
Physics news on Phys.org
Hemolymph said:

Homework Statement



from [-2,2] rad(4-x^2)

Homework Equations





The Attempt at a Solution


I know its a circle and i get the equation to be y^2+x^2=4

and I believe it has to be divided into a circle and rectangle
so the area of the rectangle i got to be 2
the circle i got to be 1/2(since its a half circle) times ∏(2)^2
the answer i got is
2+2pi (which is wrong) don't know where I went wrong tho.

Where did the rectangle come from? It's just a semicircular area.
 
Curious3141 said:
Where did the rectangle come from? It's just a semicircular area.

I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx
And the solution had a the area broken up into a rectangle and a semicircle
I guess i tried to apply the same technique to this problem.
The answer came out to be 5+(25pi)/4
 
Hemolymph said:
I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx
The 1+ gave the rectangle there. You have no corresponding term here. Did you sketch the curve? Do you see a rectangle when you do?
 
Hemolymph said:
I was doing a simliar problem of [-5,0] where it was evaluating 1+rad(25-x^2) dx
And the solution had a the area broken up into a rectangle and a semicircle
When you're using what you found in a "similar" problem, make sure it's actually similar to the one you're working on.

As suggested by others in this thread, a quick sketch of the graph of x2 + y2 = 4 would show that your region is just the upper half of a circle.

Sketching a graph is usually the first thing you need to do in these problems.
Hemolymph said:
I guess i tried to apply the same technique to this problem.
The answer came out to be 5+(25pi)/4
 
Thanks for the advice/ help
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top