Answer :
To solve the definite integral [tex]\(\int_{-9}^0 \sqrt{81-x^2} \, dx \)[/tex] using geometry, we first interpret the integrand [tex]\(\sqrt{81 - x^2}\)[/tex] geometrically.
### Step-by-Step Solution:
1. Interpret the Integrand:
The function [tex]\(\sqrt{81 - x^2}\)[/tex] represents the upper half of a circle centered at the origin with a radius of 9. To see this, note that the equation of a circle centered at the origin with radius [tex]\(r\)[/tex] is given by [tex]\(x^2 + y^2 = r^2\)[/tex]. In this case, [tex]\(r = 9\)[/tex], so the equation becomes [tex]\(x^2 + y^2 = 81\)[/tex]. Solving for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], we get [tex]\(y = \sqrt{81 - x^2}\)[/tex], which is just the upper semicircle.
2. Sketch the Graph:
- Draw a coordinate system.
- Sketch a circle centered at the origin [tex]\((0,0)\)[/tex] with a radius of 9.
- Highlight the upper semicircle described by [tex]\(y = \sqrt{81 - x^2}\)[/tex].
Now, since we are dealing with the definite integral from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex], we look at the upper semicircle from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex].
3. Identify the Region:
The integral [tex]\(\int_{-9}^0 \sqrt{81 - x^2} \, dx \)[/tex] represents the area under the curve [tex]\(y = \sqrt{81 - x^2}\)[/tex] from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex]. This is essentially the area of one-quarter of the entire circle.
4. Geometric Calculation:
- The area of a full circle with radius 9 is given by [tex]\(A = \pi r^2 = \pi \times 9^2 = 81\pi\)[/tex].
- The area of the upper semicircle is one-half of the total area: [tex]\(\dfrac{81\pi}{2}\)[/tex].
- The area under consideration, from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex], is one-half of the area of the semicircle, as it is only a quarter of the full circle. Therefore, the area is: [tex]\(\dfrac{1}{2} \left(\dfrac{81\pi}{2}\right) = \dfrac{81\pi}{4}\)[/tex].
5. Numeric Result:
Plug in the value of [tex]\(\pi \approx 3.141592653589793\)[/tex]:
[tex]\[ \dfrac{81 \pi}{4} \approx \dfrac{81 \times 3.141592653589793}{4} \approx 63.61725123519331 \][/tex]
### Final Answer:
- The integral [tex]\(\int_{-9}^0 \sqrt{81-x^2} \, dx\)[/tex] represents the area of one-fourth of a circle with radius 9. After evaluating using the area formula, the result is approximately [tex]\(63.61725123519331\)[/tex].
- Therefore, the correct graph that matches this interpretation is option A, which shows a semicircle from [tex]\(x = -9\)[/tex] to [tex]\(x = 9\)[/tex].
Thus, after interpreting the geometry and calculating the area, we see that the answer is:
(127.23450247038662, 63.61725123519331, 'A').
This indicates that the solution to the definite integral and the correct sketch is option A.
### Step-by-Step Solution:
1. Interpret the Integrand:
The function [tex]\(\sqrt{81 - x^2}\)[/tex] represents the upper half of a circle centered at the origin with a radius of 9. To see this, note that the equation of a circle centered at the origin with radius [tex]\(r\)[/tex] is given by [tex]\(x^2 + y^2 = r^2\)[/tex]. In this case, [tex]\(r = 9\)[/tex], so the equation becomes [tex]\(x^2 + y^2 = 81\)[/tex]. Solving for [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex], we get [tex]\(y = \sqrt{81 - x^2}\)[/tex], which is just the upper semicircle.
2. Sketch the Graph:
- Draw a coordinate system.
- Sketch a circle centered at the origin [tex]\((0,0)\)[/tex] with a radius of 9.
- Highlight the upper semicircle described by [tex]\(y = \sqrt{81 - x^2}\)[/tex].
Now, since we are dealing with the definite integral from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex], we look at the upper semicircle from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex].
3. Identify the Region:
The integral [tex]\(\int_{-9}^0 \sqrt{81 - x^2} \, dx \)[/tex] represents the area under the curve [tex]\(y = \sqrt{81 - x^2}\)[/tex] from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex]. This is essentially the area of one-quarter of the entire circle.
4. Geometric Calculation:
- The area of a full circle with radius 9 is given by [tex]\(A = \pi r^2 = \pi \times 9^2 = 81\pi\)[/tex].
- The area of the upper semicircle is one-half of the total area: [tex]\(\dfrac{81\pi}{2}\)[/tex].
- The area under consideration, from [tex]\(x = -9\)[/tex] to [tex]\(x = 0\)[/tex], is one-half of the area of the semicircle, as it is only a quarter of the full circle. Therefore, the area is: [tex]\(\dfrac{1}{2} \left(\dfrac{81\pi}{2}\right) = \dfrac{81\pi}{4}\)[/tex].
5. Numeric Result:
Plug in the value of [tex]\(\pi \approx 3.141592653589793\)[/tex]:
[tex]\[ \dfrac{81 \pi}{4} \approx \dfrac{81 \times 3.141592653589793}{4} \approx 63.61725123519331 \][/tex]
### Final Answer:
- The integral [tex]\(\int_{-9}^0 \sqrt{81-x^2} \, dx\)[/tex] represents the area of one-fourth of a circle with radius 9. After evaluating using the area formula, the result is approximately [tex]\(63.61725123519331\)[/tex].
- Therefore, the correct graph that matches this interpretation is option A, which shows a semicircle from [tex]\(x = -9\)[/tex] to [tex]\(x = 9\)[/tex].
Thus, after interpreting the geometry and calculating the area, we see that the answer is:
(127.23450247038662, 63.61725123519331, 'A').
This indicates that the solution to the definite integral and the correct sketch is option A.
