Find the area bounded by the y-axis and x = 4-y^2/3
Thanks!
did this one yesterday
http://www.jiskha.com/display.cgi?id=1438682725
To find the area bounded by the y-axis and the curve x = 4 - y^(2/3), we need to set up and integrate an appropriate integral.
Step 1: Determine the limits of integration
Since we are interested in the area bounded by the y-axis, we need to find the values of y where the curve intersects the y-axis (i.e., where x = 0).
0 = 4 - y^(2/3)
Rearranging the equation, we get:
y^(2/3) = 4
Taking the cube of both sides, we have:
y^2 = 64
Taking the square root of both sides, we get:
y = ±8
So, the limits of integration for the y-axis will be -8 to 8.
Step 2: Set up the integral
To find the area, we need to integrate the function x = 4 - y^(2/3) with respect to y. Since we're working with respect to y, we need to express x in terms of y.
Since x = 4 - y^(2/3), we can rewrite the integral as:
∫[from -8 to 8] (4 - y^(2/3)) dy
Step 3: Evaluate the integral
Integrate the function (4 - y^(2/3)) with respect to y from -8 to 8:
∫[from -8 to 8] (4 - y^(2/3)) dy = [4y - (3/5)y^(5/3)] evaluated from -8 to 8
Evaluating this integral, we get:
[4(8) - (3/5)(8^(5/3))] - [4(-8) - (3/5)(-8^(5/3))]
Simplifying the equation, we will get the area bounded by the y-axis and the curve x = 4 - y^(2/3).