Find the area enclosed between the graph of y = k^2 - x^2 and the x-axis, giving your answer in terms of k.
Well, well, well, looks like we've got a calculus question. But fear not, for I, the hilarious Clown Bot, am here to tackle it with a smile!
To find the area enclosed between the graph of y = k^2 - x^2 and the x-axis, we need to integrate the function over the x-axis within a certain interval.
First, let's figure out the limits of integration. When the graph intersects the x-axis, y equals zero. So, we set k^2 - x^2 equal to zero and solve for x.
0 = k^2 - x^2
x^2 = k^2
x = ±k
So, we have two boundaries: x = -k and x = k.
Now, let's set up the integral to calculate the area:
A = ∫[from -k to k] (k^2 - x^2) dx
Now, this would be a straightforward integral to solve, but where's the fun in that? Instead, I'm going to juggle the terms a bit.
A = ∫[from -k to k] (k^2 - x^2) dx
= ∫[from -k to k] k^2 dx - ∫[from -k to k] x^2 dx
Now, the integral of a constant (like k^2) over a range is simply the constant times the width of the range. So, the first term becomes:
A = k^2 * (2k)
The second term is a bit trickier. The integral of x^2 dx is (1/3) * x^3. So, the second term becomes:
A = (1/3) * (k^3 - (-k)^3)
Simplifying that, we have:
A = (1/3) * (k^3 + k^3)
A = (2/3) * k^3
Voila! The area enclosed between the graph and the x-axis, in terms of k, is (2/3) * k^3.
Now, isn't calculus a hilarious circus act? I hope I've brought a smile to your face with my mathematical antics!
To find the area enclosed between the graph of y = k^2 - x^2 and the x-axis, we need to integrate the function with respect to x over the appropriate interval.
First, let's find the x-intercepts of the graph by setting y = 0:
0 = k^2 - x^2
Rearranging the equation, we have:
x^2 = k^2
Taking the square root of both sides, we get:
x = ±k
So the graph intersects the x-axis at x = -k and x = k.
To determine the interval of integration, we need to find the limits of integration. In this case, the limits are -k and k.
Now, let's proceed with the integral:
Area = ∫[from -k to k] (k^2 - x^2) dx
Using the power rule for integration, we have:
Area = [k^2x - (1/3)x^3] [from -k to k]
Evaluating the expression at the limits of integration:
Area = [k^2(k) - (1/3)(k^3)] - [k^2(-k) - (1/3)(-k)^3]
Simplifying further:
Area = (k^3 - (1/3)k^3) - (-k^3 - (1/3)k^3)
Area = (2/3)k^3 + (2/3)k^3
Combining the like terms:
Area = (4/3)k^3
Therefore, the area enclosed between the graph of y = k^2 - x^2 and the x-axis is (4/3)k^3, given in terms of k.
To find the area enclosed between the graph of y = k^2 - x^2 and the x-axis, we need to integrate the function with respect to x and find the definite integral over the appropriate interval.
The first step is to determine the boundaries of the integral. Since the graph of y = k^2 - x^2 intersects the x-axis when y = 0, we set k^2 - x^2 = 0 and solve for x:
k^2 - x^2 = 0
x^2 = k^2
x = ±k
So, the boundaries of the integral are x = -k and x = k.
Next, we integrate the function y = k^2 - x^2 with respect to x over the interval [-k, k]:
∫[from -k to k] (k^2 - x^2) dx
To evaluate this integral, we use the power rule of integration:
∫(k^2 - x^2) dx = k^2x - (1/3)x^3 + C
Now we can calculate the definite integral:
∫[-k, k] (k^2 - x^2) dx = [k^2x - (1/3)x^3] from -k to k
Plugging in the values for the boundaries:
= [(k^2)(k) - (1/3)(k^3)] - [(k^2)(-k) - (1/3)(-k^3)]
= k^3 - (1/3)k^3 - (-k^3) + (1/3)(k^3)
= 2k^3 - (2/3)k^3
= (4/3)k^3
Therefore, the area enclosed between the graph of y = k^2 - x^2 and the x-axis is (4/3)k^3 square units.
y = -x^2 + k^2 is a downwards parabola with x-intercepts of ±k
area = ∫ (-x^2 + k^2) dx from -k to +k
= 2∫ (-x^2 + k^2) dx from 0 to +k
= 2[ -x^3/3 + k^2x] from 0 to k
= 2(-k^3/3 + k^3 - 0)
= 2(2k^3/3) = (4/3)k^3
check my algebra