Question

(1 point) find the volume formed by rotating the region enclosed by: ( x = 9y ) and ( y^3 = x ) with ( y geq 0 ) about the y-axis. volume = preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining.

Explanation:

Step1: Find Intersection Points

Set \( 9y = y^3 \) (since \( y\geq0 \), we can solve for \( y \)).
\( y^3 - 9y = 0 \)
\( y(y^2 - 9)=0 \)
\( y(y - 3)(y + 3)=0 \)
Since \( y\geq0 \), the solutions are \( y = 0 \) and \( y = 3 \).

Step2: Set Up the Integral for Volume (Washer Method)

The outer radius \( R(y) \) and inner radius \( r(y) \) when rotating about the \( y \)-axis:
For a given \( y \), \( x = 9y \) and \( x = y^3 \). We need to determine which is larger in \( [0, 3] \).
At \( y = 1 \), \( 9(1)=9 \) and \( 1^3 = 1 \), so \( 9y\geq y^3 \) on \( [0, 3] \).
So \( R(y)=9y \), \( r(y)=y^3 \).
The formula for the volume using the washer method about the \( y \)-axis is \( V=\pi\int_{a}^{b}[R(y)^2 - r(y)^2]dy \)
Here, \( a = 0 \), \( b = 3 \), so:
\( V=\pi\int_{0}^{3}[(9y)^2-(y^3)^2]dy \)
\( V=\pi\int_{0}^{3}(81y^2 - y^6)dy \)

Step3: Integrate the Function

Integrate term by term:
\( \int 81y^2 dy=81\times\frac{y^3}{3}=27y^3 \)
\( \int y^6 dy=\frac{y^7}{7} \)
Evaluate from \( 0 \) to \( 3 \):
\( [27y^3-\frac{y^7}{7}]_{0}^{3}=27(3)^3-\frac{3^7}{7}-(0 - 0) \)
\( 27\times27-\frac{2187}{7}=\frac{729\times7 - 2187}{7}=\frac{5103 - 2187}{7}=\frac{2916}{7} \)

Step4: Multiply by \( \pi \)

\( V=\pi\times\frac{2916}{7}=\frac{2916\pi}{7} \)

Answer:

\(\frac{2916\pi}{7}\)