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the area of region enclosed by the curve y=x^2-4x+4 and y^2=16-8x


Lalitha P 11th Feb, 2025
Answer (1)
Poral Siddartha (siddu) 14th Feb, 2025

We need to find the area enclosed by the two curves:

1. Parabola: y = x² 4x + 4

2. Sideways parabola: y² = 16 - 8x 2

Step 1: Find Intersection Points

The enclosed area is where these two curves cross each other. To find this, we substitute the equation for y from the first curve into the second and solve for x. This gives us the limits for integration.

Step 2: Set Up the Integral

Once we know the intersection points, we determine which curve is on top (higher value of y) and which is at the bottom. The area is then found by integrating the difference between them.

To find the area of the region enclosed by the curves y = x ^ 2 - 4x + 4 and y ^ 2 = 16 - 8x . follow these steps:

1. Find the points of intersection: Solve y = x ^ 2 - 4x + 4 and y ^ 2 = 16 - 8x

Substitute y = x ^ 2 - 4x + 4 into y ^ 2 = 16 - 8x

(x ^ 2 - 4x + 4) ^ 2 = 16 - 8x

1. Set up the integral: After finding the intersection points, use definite integration to compute the area between the curves.

2. Calculate the area:

Area = integrate |y_{2}(x) - y_{1}(x)| dx from x_{1} to x_{2}

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