Find the volume formed by the revolution about the y-axis of the part of te parabola y^2=4ax cut off by it's latus rectum.

To find the volume formed by revolving the part of the parabola \(y^2 = 4ax\) about the y-axis, specifically the section cut off by its latus rectum, we first need to understand a few key concepts about the geometry of the parabola and the method of calculating volumes of revolution.

Understanding the Parabola

The equation \(y^2 = 4ax\) describes a rightward-opening parabola. The parameter \(a\) represents the distance from the vertex of the parabola to its focus, which is located at the point \((a, 0)\). The latus rectum of this parabola is a line segment that passes through the focus and is perpendicular to the axis of symmetry of the parabola. Its endpoints lie on the parabola itself.

Finding the Latus Rectum

The length of the latus rectum for this parabola is \(4a\). The endpoints of the latus rectum can be found by substituting \(x = a\) into the parabola's equation:

• Substituting \(x = a\) gives \(y^2 = 4a(a) = 4a^2\), leading to \(y = \pm 2a\).

Thus, the endpoints of the latus rectum are \((a, 2a)\) and \((a, -2a)\).

Setting Up the Volume Integral

To find the volume of the solid formed by revolving this section of the parabola around the y-axis, we can use the method of cylindrical shells. The formula for the volume \(V\) using this method is:

V = 2π ∫ (radius)(height) dy

In our case, the radius is the horizontal distance from the y-axis to the parabola, which can be expressed in terms of \(y\). From the equation \(y^2 = 4ax\), we can solve for \(x\):

• x = (y^2)/(4a)

Determining the Integration Limits

Next, we need to determine the limits of integration. Since we are interested in the portion of the parabola cut off by the latus rectum, the limits will be from \(-2a\) to \(2a\) (the y-coordinates of the endpoints of the latus rectum).

Calculating the Volume

Now, we can set up our integral:

V = 2π ∫ from -2a to 2a of (y^2/(4a))(y) dy

This simplifies to:

V = 2π/(4a) ∫ from -2a to 2a of y^3 dy

We can compute the integral:

∫ y^3 dy = (1/4)y^4, so:

∫ from -2a to 2a of y^3 dy = [(1/4)(2a)^4] - [(1/4)(-2a)^4] = (1/4)(16a^4) - (1/4)(16a^4) = 0

This indicates that the contributions from both sides of the y-axis cancel out when integrated over a symmetric interval around zero.

Finalizing the Volume Calculation

However, since we need the absolute value for the volume and considering the symmetry, we can compute the volume from 0 to \(2a\) and double it:

V = 2 * (2π/(4a) * ∫ from 0 to 2a of y^3 dy)

Calculating this integral gives:

∫ from 0 to 2a of y^3 dy = (1/4)(2a)^4 = 4a^4

Thus, the volume becomes:

V = 2 * (2π/(4a) * 4a^4) = 4πa^3

Final Result

The total volume of the solid formed by revolving the part of the parabola \(y^2 = 4ax\) cut off by its latus rectum about the y-axis is: