Compute the area of a leaf given by the equation y^2 = (x^2−1)^2 for −1 ≤ x ≤ 1
Using fundamental theorem of calculus, i got the answer -4/3 but on the answer key it isn't.
I get for the area
∫[-1,1] (1-x^2)-(x^2-1) dx = ∫[-1,1] 2-2x^2 dx = 2(x - x^3/3)[-1,1] = 8/3
can you explain it please? where did (1-x^2)-(x^2-1) come from?
y^2 = (x^2−1)^2
y = ±(x^2-1)
That is, y = (x^2-1) or y = (1-x^2)
The area between two functions f(x) and g(x) is ∫ f(x)-g(x) dx.
hat is, it is the sum of all the thin strips whose height is the difference between the curves.
To compute the area of the leaf, we can use the concept of integration in calculus. First, let's draw the graph of the given equation to visualize the shape of the leaf.
The equation y^2 = (x^2 - 1)^2 can be simplified as y = x^2 - 1 and y = -(x^2 - 1). Here's how the graph looks like:
y |
|
| x
|_______
_|_
The leaf has two halves: the upper half and the lower half. We can find the area of each half separately and then sum them together to get the total area.
Let's start by finding the area of the upper half. To do this, we need to find the limits of integration. Since the leaf is symmetric about the y-axis, we can integrate from x = -1 to x = 1 and multiply the result by 2.
The formula to find the area under the curve is:
A = ∫[a,b] f(x) dx
For the upper half, we have f(x) = x^2 - 1.
Now, let's integrate the function f(x) from x = -1 to x = 1:
A_upper = 2 * ∫[-1,1] (x^2 - 1) dx
Evaluating the integral:
A_upper = 2 * [ (1/3)x^3 - x ] | [-1,1]
Plugging in the values:
A_upper = 2 * ( (1/3)(1)^3 - 1 ) - ( (1/3)(-1)^3 - (-1) )
= 2 * (1/3 - 1) - (1/3 + 1)
= 2 * (-2/3) - (4/3)
= -4/3 - 4/3
= -8/3
The area of the upper half is -8/3 square units.
Similarly, we can find the area of the lower half. Again, because of symmetry, the area will be the same as the upper half. So, A_lower = -8/3 square units.
Therefore, the total area of the leaf is:
A = A_upper + A_lower = (-8/3) + (-8/3) = -16/3.
Hence, the area of the leaf given by the equation y^2 = (x^2 - 1)^2 for -1 ≤ x ≤ 1 is -16/3 square units.