Solve Solid of Revolution Homework: Area Enclosed by Ellipse & Auxiliary Circle

[Parthalan]

Homework Statement

The area enclosed between the ellipse $4x^2 + 9y^2 = 36$ and its auxiliary circle $x^2 + y^2 = 9$ is rotated about the y-axis through $\pi$ radians. Find, by integration, the volume generated.

This is the whole question. I assume it means bounded by the x-axis, but even if this isn't the case, my answer is wrong. :(

Solution: $12\pi$

Homework Equations

$$V = \int_a^b{A(y)\,dy}$$

The Attempt at a Solution

$$x^2 + y^2 = 9 \Rightarrow x^2 = 9 - y^2$$

$$4x^2 + 9y^2 = 36 \Rightarrow x^2 = \frac{36-9y^2}{4}$$

$$A(y) = \pi \left [ \left ( 9-y^2 \right ) - \left ( \frac{36-9y^2}{4} \right ) \right ] = \frac{5\pi}{4}y^2$$

$$\int_0^3{\frac{5\pi}{4}y^2\,dy} = \frac{5\pi}{4} \left [ \frac{y^3}{3} \right ]^3_0 = \frac{45\pi}{4}$$

I hope someone can help me find where I went wrong! Thanks very much.

 

[HallsofIvy]

I see no reason to assume "bounded" by the y-axis. There are "natural" bounds at x= -3 and x= 3 and if nothing else is said, I think those should be used.

Of course, x²+ y²= 9 is a cirlcle with center at (0,0) and radius 3. Rotated around any axis it makes a sphere with center at (0,0,0) and radius 3. Its volume is $(4/3)\pi 3^3= 36\pi$.

$4x^2+ 9y^2= 36$ is an ellipse with center at (0,0), semi-major axis along the x-axisd of length 3 and semi-minor axis along the y-axis of length 2. Rotating around the y-axis makes an ellipsoid with center at (0,0,0) and semi-axes of length 2, 3 and 3. Its volume is given by $(4/3)\pi (3)(3)(2)= 24\pi$.

The volume between the two figures is $36\pi- 24\pi= 12\pi$. However, if it is only rotated by $\pi$ radians, that is only half a circle so we should ony get 1/2 the volume: $6\pi$. Are you sure about that? Of course, that's not "by integration" but serves as a check.

The problem is that the ellipse does not extend above y= 2. I think you need to do this in parts: for 0< y< 2, you do, in fact, have $\pi [(9- x^2)- (36- 9x^2)/4]= (5/4)\pi x^2$. But for 2< y< 3, you have the full $\pi(9- x^2)$.

 

[Parthalan]

I *think* I get it now. Because $4x^2+9y^2=36$ doesn't extend above $y=2$, I needed to "partition" the volume and evaluate
$$\pi \left ( \frac{5}{4}\int_0^2{x^2\,dx} + \int_2^3{9-x^2\,dx} \right ) = 6\pi$$
and then either double for the part below the x-axis, or work it out by hand using the same idea.

Thanks very much!