Area bounded by the y-axis and the curve x = y² - y³

[tandoorichicken]

Find the area bounded by the y-axis and the curve x = y^2 - y^3
What exactly does the graph of x = y^2 - y^3 look like? And how do you set up the integration?

 

[ShawnD]

Originally posted by tandoorichicken
Find the area bounded by the y-axis and the curve x = y^2 - y^3
What exactly does the graph of x = y^2 - y^3 look like? And how do you set up the integration?

Interchange the x and y, does it look familiar now? Rotate that by 90 degrees and you'll get the picture.

intercepts:

$$y^2 - y^3 = 0$$

$$y^2(1 - y) = 0$$

y = 0, y = 1

integration:

$$A = \int^1_0 y^2 - y^3 dy$$

$$A = \frac{y^3}{3} - \frac{y^4}{4} |^1_0$$

$$A = \frac{1^3}{3} - \frac{1^4}{4}$$

$$A = \frac{1}{3} - \frac{1}{4}$$

$$A = \frac{1}{12}$$

That's my answer anyway.

 

[M98Ranger]

The graph of x = y^2 - y^3 is a parabola that opens to the right and has a vertex at (0,0). As y increases, the parabola decreases in height until it reaches the x-axis at y=1, and then it continues to decrease as it approaches the y-axis.

To set up the integration, we can use the fact that the area under a curve is given by the definite integral of the function. In this case, since we are looking for the area bounded by the y-axis and the curve, we can integrate from y=0 to y=1 (the y-values where the curve intersects the y-axis). This will give us the area between the y-axis and the curve.

The integral would be ∫(y^2-y^3)dy from y=0 to y=1. We can then use the power rule to solve the integral, which would result in the area being equal to 1/12 square units. Graphically, this would look like the area under the curve between the y-axis and the curve itself.