Answer :
To find the area bounded by the curve [tex]\( y^2 = x^2 \)[/tex] and the line [tex]\( x = 8 \)[/tex], let's break the problem down step-by-step:
1. Understanding the Curve:
- The equation [tex]\( y^2 = x^2 \)[/tex] can be simplified to [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex]. These represent two lines: one is the line [tex]\( y = x \)[/tex], and the other is the line [tex]\( y = -x \)[/tex].
2. Region of Interest:
- We are interested in the region bounded by these lines up to [tex]\( x = 8 \)[/tex]. Specifically, we need to find the area between the curves [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex].
3. Setting Up the Integral:
- To calculate the area, we'll consider the area between [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex].
- The lines [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] intersect the x-axis symmetrically, effectively stretching from [tex]\(-8\)[/tex] to [tex]\(8\)[/tex], but we only need to consider the positive x-axis due to the symmetry.
4. Calculating the Area:
- The absolute value function [tex]\( |x| \)[/tex] represents the distance from the x-axis that both lines [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] cover.
- We sum the absolute value function from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex] to find the enclosed area: [tex]\(\int_{0}^{8} 2|x| \, dx\)[/tex].
- Since [tex]\( x \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex] within positive values, [tex]\(|x| = x\)[/tex]. Thus, the integral becomes [tex]\(\int_{0}^{8} 2x \, dx\)[/tex].
5. Evaluating the Integral:
- To integrate [tex]\(\int_{0}^{8} 2x \, dx\)[/tex], we find the antiderivative of [tex]\( 2x \)[/tex], which is [tex]\( x^2 \)[/tex].
- Evaluate this antiderivative from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex]: [tex]\([x^2]_{0}^{8} = 8^2 - 0^2 = 64\)[/tex].
6. Conclusion:
- Hence, the area bounded by the curve [tex]\( y^2 = x^2 \)[/tex] and the line [tex]\( x=8 \)[/tex] is [tex]\( 64 \)[/tex] square units.
Therefore, the correct option is:
(B) 64 sq. units.
1. Understanding the Curve:
- The equation [tex]\( y^2 = x^2 \)[/tex] can be simplified to [tex]\( y = x \)[/tex] or [tex]\( y = -x \)[/tex]. These represent two lines: one is the line [tex]\( y = x \)[/tex], and the other is the line [tex]\( y = -x \)[/tex].
2. Region of Interest:
- We are interested in the region bounded by these lines up to [tex]\( x = 8 \)[/tex]. Specifically, we need to find the area between the curves [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex].
3. Setting Up the Integral:
- To calculate the area, we'll consider the area between [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] from [tex]\( x = 0 \)[/tex] to [tex]\( x = 8 \)[/tex].
- The lines [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] intersect the x-axis symmetrically, effectively stretching from [tex]\(-8\)[/tex] to [tex]\(8\)[/tex], but we only need to consider the positive x-axis due to the symmetry.
4. Calculating the Area:
- The absolute value function [tex]\( |x| \)[/tex] represents the distance from the x-axis that both lines [tex]\( y = x \)[/tex] and [tex]\( y = -x \)[/tex] cover.
- We sum the absolute value function from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex] to find the enclosed area: [tex]\(\int_{0}^{8} 2|x| \, dx\)[/tex].
- Since [tex]\( x \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex] within positive values, [tex]\(|x| = x\)[/tex]. Thus, the integral becomes [tex]\(\int_{0}^{8} 2x \, dx\)[/tex].
5. Evaluating the Integral:
- To integrate [tex]\(\int_{0}^{8} 2x \, dx\)[/tex], we find the antiderivative of [tex]\( 2x \)[/tex], which is [tex]\( x^2 \)[/tex].
- Evaluate this antiderivative from [tex]\( 0 \)[/tex] to [tex]\( 8 \)[/tex]: [tex]\([x^2]_{0}^{8} = 8^2 - 0^2 = 64\)[/tex].
6. Conclusion:
- Hence, the area bounded by the curve [tex]\( y^2 = x^2 \)[/tex] and the line [tex]\( x=8 \)[/tex] is [tex]\( 64 \)[/tex] square units.
Therefore, the correct option is:
(B) 64 sq. units.