Find the area of the bounded region of
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
Step 1: Recognize that this is the equation of an ellipse in standard form: \[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \] where \( a = 4 \) and \( b = 3 \) are the semi-major and semi-minor axes, respectively.
Step 2: The area of an ellipse is given by the formula: \[ A = \pi \cdot a \cdot b \] Substitute the values of \( a \) and \( b \): \[ A = \pi \cdot 4 \cdot 3 = 12\pi. \] Thus, the area of the bounded region is \( 12\pi \).
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Find the values of \( x, y, z \) if the matrix \( A \) satisfies the equation \( A^T A = I \), where
\[ A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} \]