Answer:
The integral of the volume is:
[tex]V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy[/tex]
The result is: [tex]V = 78.97731[/tex]
Step-by-step explanation:
Given
Curve: [tex]x^2 + 4y^2 = 4[/tex]
About line [tex]x = 2[/tex] --- Missing information
Required
Set up an integral for the volume
[tex]x^2 + 4y^2 = 4[/tex]
Make x^2 the subject
[tex]x^2 = 4 - 4y^2[/tex]
Square both sides
[tex]x = \sqrt{(4 - 4y^2)[/tex]
Factor out 4
[tex]x = \sqrt{4(1 - y^2)[/tex]
Split
[tex]x = \sqrt{4} * \sqrt{(1 - y^2)[/tex]
[tex]x = \±2 * \sqrt{(1 - y^2)[/tex]
[tex]x = \±2 \sqrt{(1 - y^2)[/tex]
Split
[tex]x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}[/tex]
Rotate about x = 2 implies that:
[tex]r = 2 - x[/tex]
So:
[tex]r_1 = 2 - (-2 \sqrt{(1 - y^2)})[/tex]
[tex]r_1 = 2 +2 \sqrt{(1 - y^2)}[/tex]
[tex]r_2 = 2 - 2 \sqrt{(1 - y^2)}[/tex]
Using washer method along the y-axis i.e. integral from 0 to 1.
We have:
[tex]V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy[/tex]
Substitute values for r1 and r2
[tex]V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy[/tex]
Evaluate the squares
[tex]V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy[/tex]
Remove brackets and collect like terms
[tex]V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2) - 4(1 - y^2)} \, dy[/tex]
[tex]V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy[/tex]
Rewrite as:
[tex]V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy[/tex]
[tex]V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy[/tex]
Using the calculator:
[tex]\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}[/tex]
So:
[tex]V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy[/tex]
[tex]V = 32\pi * \frac{\pi}{4}[/tex]
[tex]V =\frac{32\pi^2}{4}[/tex]
[tex]V =8\pi^2[/tex]
Take:
[tex]\pi = 3.142[/tex]
[tex]V = 8* 3.142^2[/tex]
[tex]V = 78.97731[/tex] --- approximated
