Answer
a.) $V=78.95684$
b.) $V=78.95684$
Work Step by Step
a.)
$x^2+4y^2=4\\
x^2+4(0)^2=4\\
x^2=4\\
x=-2 \qquad x=2$
$\displaystyle{A(x)=\pi\left(2+\sqrt{\frac{4-x^2}{4}}\right)^2-\pi\left(2-\sqrt{\frac{4-x^2}{4}}\right)^2}\\
\displaystyle{A(x)=\pi\left(4\sqrt{4-x^2}\right)}$
$\begin{aligned} V &=\int_{-2}^{2} A(x) \ d x \\ V &=\int_{-2}^{2} \pi\left(4\sqrt{4-x^2}\right) \ d x \\ V &=4\pi \int_{-2}^{2} \sqrt{4-x^2}\ dx
\\ V &=8\pi \int_{0}^{2} \sqrt{4-x^2}\ dx \\
V &=78.95684 \end{aligned}$
b.)
$x^2+4y^2=4\\
(0)^2+4y^2=4\\
y^2=1\\
y=-1 \qquad y=1$
$\displaystyle{A(y)=\pi\left(2+\sqrt{4-4y^2}\right)^2-\pi\left(2-\sqrt{4-4y^2}\right)^2}\\
\displaystyle{A(y)=\pi\left(8\sqrt{4-4y^2}\right)}$
$\begin{aligned} V &=\int_{-1}^{1} A(y) \ d y \\ V &=\int_{-1}^{1} \pi\left(8\sqrt{4-4y^2}\right) \ d y \\ V &=8\pi \int_{-1}^{1} \sqrt{4-4y^2}\ dy
\\ V &=16\pi \int_{0}^{1} \sqrt{4-4y^2}\ dy \\
V &=78.95684 \end{aligned}$