- #1
MarkFL
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MHB
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Here is the question:
I have posted a link there to this topic so the OP can see my work.
I have posted a link there to this topic so the OP can see my work.
Find Area Between Circle & Function: Calc II
In summary, RS MASTER was asking for the area bounded by a function and a circle with a radius of 1. He found the area using the FTOC and then used geometry to confirm the result.
- #2
MarkFL
Gold Member
MHB
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Re: RS MASTER's question at Yahoo! Answers regading finding the area bounded by a function and circl
Hello RS MASTER,
I think what I would do, is center the circle at the origin of a new coordinate system, and then let the first quadrant line be:
\(\displaystyle y=-\frac{1}{2}x+b\) where $1\le b$
Now, we may determine the parameter $b$, by substituting for $y$ in the equation of the circle, using the linear equation, and then equate the discriminant of the resulting quadratic in $x$ to zero, since a tangent line will touch the circle in one place only, and so the resulting quadratic can only have one root. Hence:
\(\displaystyle x^2+\left(-\frac{1}{2}x+b \right)^2=1\)
Writing the quadratic in standard form, we find:
\(\displaystyle x^2+\frac{1}{4}x^2-bx+b^2=1\)
\(\displaystyle \frac{5}{4}x^2-bx+b^2-1=0\)
\(\displaystyle 5x^2-4bx+4\left(b^2-1 \right)=0\)
Equating the discriminant to zero, we find:
\(\displaystyle (-4b)^2-4(5)\left(4\left(b^2-1 \right) \right)=0\)
\(\displaystyle b^2-5\left(b^2-1 \right)=0\)
\(\displaystyle b^2=\frac{5}{4}\)
Taking the positive root, we find:
\(\displaystyle b=\frac{\sqrt{5}}{2}\)
Here is a plot of the area we will find, which we will then double to answer the given question:
View attachment 1426
Next, we need to determine the upper limit of integration, which is the $x$-coordinate of the point of tangency between the line and the circle. Thus, we need to solve:
\(\displaystyle 5x^2-2\sqrt{5}x+1=0\)
\(\displaystyle \left(\sqrt{5}x-1 \right)^2=0\)
\(\displaystyle x=\frac{1}{\sqrt{5}}\)
And so, the area $A$ we seek is given by:
\(\displaystyle A=2\int_0^{\frac{1}{\sqrt{5}}} \frac{\sqrt{5}-x}{2}-\sqrt{1-x^2}\,dx=\int_0^{\frac{1}{\sqrt{5}}} \sqrt{5}-x-2\sqrt{1-x^2}\,dx\)
At this point, it will be useful to develop a formula to handle integrals of the form:
\(\displaystyle I=\int\sqrt{a^2-x^2}\,dx\)
Let's try the trigonometric substitution:
\(\displaystyle x=a\sin(\theta)\,\therefore\,dx=a\cos(\theta)\, d\theta\)
\(\displaystyle I=a^2\int\cos^2(\theta)\,d\theta\)
Using a double-angle identity for cosine, we may write:
\(\displaystyle I=\frac{a^2}{2}\int 1+\cos(2\theta)\,d\theta\)
\(\displaystyle I=\frac{a^2}{2}\left( \theta+\frac{1}{2}\sin(2 \theta)+C \right)\)
Using the double-angle identity for sine, we have:
\(\displaystyle I=\frac{a^2}{2}\left( \theta+\sin( \theta)\cos( \theta)+C \right)\)
Back-substituting for $\theta$, we obtain:
\(\displaystyle I=\frac{a^2}{2}\left(\sin^{-1}\left(\frac{x}{a} \right)+\frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}+C \right)\)
\(\displaystyle I=\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a} \right)+\frac{x}{2}\sqrt{a^2-x^2}+C\)
Now, back to the integral representing the area $A$, we may now apply the FTOC as follows:
\(\displaystyle A=\left[\sqrt{5}x-\frac{1}{2}x^2-\sin^{-1}(x)-x\sqrt{1-x^2} \right]_0^{\frac{1}{\sqrt{5}}}\)
\(\displaystyle A=1-\frac{1}{10}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)-\frac{2}{5}=\frac{1}{2}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
Now, we may check our result using some geometry. If we draw a radius of the circle from the origin to the point of tangency, we find the angle subtended by the $y$-axis and this radius is:
\(\displaystyle \alpha=\frac{\pi}{2}-\cos^{-1}\left(\frac{1}{\sqrt{5}} \right)=\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
Now, using the formula for the area of a triangle, and the formula for the area of a circular sector, we may write:
\(\displaystyle A=2\left(\frac{1}{2}(1)\left(\frac{\sqrt{5}}{2} \right)\sin\left(\sin^{-1}\left(\frac{1}{\sqrt{5}} \right) \right)-\frac{1}{2}(1)^2\sin^{-1}\left(\frac{1}{\sqrt{5}} \right) \right)\)
\(\displaystyle A=\frac{1}{2}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
And this checks with our result obtained from the calculus. (Sun)
Hello RS MASTER,
I think what I would do, is center the circle at the origin of a new coordinate system, and then let the first quadrant line be:
\(\displaystyle y=-\frac{1}{2}x+b\) where $1\le b$
Now, we may determine the parameter $b$, by substituting for $y$ in the equation of the circle, using the linear equation, and then equate the discriminant of the resulting quadratic in $x$ to zero, since a tangent line will touch the circle in one place only, and so the resulting quadratic can only have one root. Hence:
\(\displaystyle x^2+\left(-\frac{1}{2}x+b \right)^2=1\)
Writing the quadratic in standard form, we find:
\(\displaystyle x^2+\frac{1}{4}x^2-bx+b^2=1\)
\(\displaystyle \frac{5}{4}x^2-bx+b^2-1=0\)
\(\displaystyle 5x^2-4bx+4\left(b^2-1 \right)=0\)
Equating the discriminant to zero, we find:
\(\displaystyle (-4b)^2-4(5)\left(4\left(b^2-1 \right) \right)=0\)
\(\displaystyle b^2-5\left(b^2-1 \right)=0\)
\(\displaystyle b^2=\frac{5}{4}\)
Taking the positive root, we find:
\(\displaystyle b=\frac{\sqrt{5}}{2}\)
Here is a plot of the area we will find, which we will then double to answer the given question:
View attachment 1426
Next, we need to determine the upper limit of integration, which is the $x$-coordinate of the point of tangency between the line and the circle. Thus, we need to solve:
\(\displaystyle 5x^2-2\sqrt{5}x+1=0\)
\(\displaystyle \left(\sqrt{5}x-1 \right)^2=0\)
\(\displaystyle x=\frac{1}{\sqrt{5}}\)
And so, the area $A$ we seek is given by:
\(\displaystyle A=2\int_0^{\frac{1}{\sqrt{5}}} \frac{\sqrt{5}-x}{2}-\sqrt{1-x^2}\,dx=\int_0^{\frac{1}{\sqrt{5}}} \sqrt{5}-x-2\sqrt{1-x^2}\,dx\)
At this point, it will be useful to develop a formula to handle integrals of the form:
\(\displaystyle I=\int\sqrt{a^2-x^2}\,dx\)
Let's try the trigonometric substitution:
\(\displaystyle x=a\sin(\theta)\,\therefore\,dx=a\cos(\theta)\, d\theta\)
\(\displaystyle I=a^2\int\cos^2(\theta)\,d\theta\)
Using a double-angle identity for cosine, we may write:
\(\displaystyle I=\frac{a^2}{2}\int 1+\cos(2\theta)\,d\theta\)
\(\displaystyle I=\frac{a^2}{2}\left( \theta+\frac{1}{2}\sin(2 \theta)+C \right)\)
Using the double-angle identity for sine, we have:
\(\displaystyle I=\frac{a^2}{2}\left( \theta+\sin( \theta)\cos( \theta)+C \right)\)
Back-substituting for $\theta$, we obtain:
\(\displaystyle I=\frac{a^2}{2}\left(\sin^{-1}\left(\frac{x}{a} \right)+\frac{x}{a}\cdot\frac{\sqrt{a^2-x^2}}{a}+C \right)\)
\(\displaystyle I=\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a} \right)+\frac{x}{2}\sqrt{a^2-x^2}+C\)
Now, back to the integral representing the area $A$, we may now apply the FTOC as follows:
\(\displaystyle A=\left[\sqrt{5}x-\frac{1}{2}x^2-\sin^{-1}(x)-x\sqrt{1-x^2} \right]_0^{\frac{1}{\sqrt{5}}}\)
\(\displaystyle A=1-\frac{1}{10}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)-\frac{2}{5}=\frac{1}{2}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
Now, we may check our result using some geometry. If we draw a radius of the circle from the origin to the point of tangency, we find the angle subtended by the $y$-axis and this radius is:
\(\displaystyle \alpha=\frac{\pi}{2}-\cos^{-1}\left(\frac{1}{\sqrt{5}} \right)=\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
Now, using the formula for the area of a triangle, and the formula for the area of a circular sector, we may write:
\(\displaystyle A=2\left(\frac{1}{2}(1)\left(\frac{\sqrt{5}}{2} \right)\sin\left(\sin^{-1}\left(\frac{1}{\sqrt{5}} \right) \right)-\frac{1}{2}(1)^2\sin^{-1}\left(\frac{1}{\sqrt{5}} \right) \right)\)
\(\displaystyle A=\frac{1}{2}-\sin^{-1}\left(\frac{1}{\sqrt{5}} \right)\)
And this checks with our result obtained from the calculus. (Sun)
Attachments
Related to Find Area Between Circle & Function: Calc II
1. What is the formula for finding the area between a circle and a function?
The formula for finding the area between a circle and a function is given by: A = ∫[b,a] (πr^2 - f(x)^2)dx, where r is the radius of the circle and f(x) is the function that defines the curve.
2. How do you determine the limits of integration for finding the area between a circle and a function?
The limits of integration for finding the area between a circle and a function are determined by finding the points of intersection between the circle and the function. These points will serve as the endpoints of the integration interval.
3. Can the area between a circle and a function be negative?
No, the area between a circle and a function cannot be negative. This is because the area is calculated using the definite integral, which always yields a positive value.
4. What is the significance of finding the area between a circle and a function?
Finding the area between a circle and a function can be useful in real-world scenarios, such as calculating the surface area of a curved object or determining the volume of a solid with a curved base. It also has applications in physics and engineering, such as calculating the work done by a force on a moving object.
5. Are there any special cases when finding the area between a circle and a function?
Yes, there are some special cases when finding the area between a circle and a function. For example, if the function lies completely inside the circle or if the function is a straight line, the area will be equal to the area of the circle. If the function intersects the circle at multiple points, the area will need to be calculated using multiple integrals.
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