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Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Then use your calculator to evaluate the integral correct to five decimal places.

$x^2 + 4y^2 = 4$

(a) About $y=2$ (b) About $x = 2$

I tried using pi(integrand)((root(1- x^2/4))^2-(2^2))dx but i don't think that's right... this is another weird question... never seen an example like it..

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Hints: Draw a picture of the ellipse, and the axes of rotation.

(a) The solid will have a hole in it. We do slicing perpendicular to the $x$-axis, and need to find the cross-sectional area of the "washer" at $x$.

The bottom half of the ellipse has equation $y=-\frac{1}{2}\sqrt{4-x^2}$, and the top half has equation $y=\frac{1}{2}\sqrt{4-x^2}$.

Thus the outer radius of the washer is $2-(-\frac{1}{2}\sqrt{4-x^2})=2+\frac{1}{2}\sqrt{4-x^2}$. Call this $R(x)$.

Find an expression for the inner radius $r(x)$.

The volume is $$\int_{-2}^2 \pi\left(R^2(x)-r^2(x)\right)\,dx.$$

You will probably want to simplify the expression before further processing.

(b) This is easiest by the method often called the Method of Cylindrical Shells.

Take a vertical slice of our ellipse, say "at" $x$ and of thickness $dx$. Rotate this about the line $x=2$. We get a thin shell of radius $2-x$, and height $2y$, where $y=\frac{1}{2}\sqrt{4-x^2}$.

Use the above information to find an approximate expression for the volume of the shell, and integrate to find the desired volume.

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