Questions: Given the following double integral in polar coordinates [ int0^pi/4 int0^9(6 r^2 sin (2 theta)) r d r d theta ] Find the value of the inside integral. (A) 0.5 cos (2 theta) (B) 9841.5 sin (2 theta) (C) 1458 sin (2 theta) (D) 4920.75 cos (2 theta)

Solution Steps

To solve the given double integral in polar coordinates, we first need to evaluate the inside integral with respect to \( r \). The inside integral is: \[ \int_{0}^{9} 6 r^{2} \sin (2 \theta) \cdot r \, dr \] This simplifies to: \[ 6 \sin (2 \theta) \int_{0}^{9} r^{3} \, dr \] We can then integrate \( r^{3} \) with respect to \( r \) from 0 to 9.

1. Extract the constant \( 6 \sin (2 \theta) \) from the integral.
2. Integrate \( r^{3} \) with respect to \( r \) from 0 to 9.
3. Multiply the result by \( 6 \sin (2 \theta) \).

We start with the inside integral of the given double integral: \[ \int_{0}^{9} 6 r^{2} \sin (2 \theta) \cdot r \, dr \] This simplifies to: \[ 6 \sin (2 \theta) \int_{0}^{9} r^{3} \, dr \]

Next, we evaluate the integral: \[ \int_{0}^{9} r^{3} \, dr = \left[ \frac{r^{4}}{4} \right]_{0}^{9} = \frac{9^{4}}{4} = \frac{6561}{4} \]

Now, we multiply the result of the integral by \( 6 \sin (2 \theta) \): \[ 6 \sin (2 \theta) \cdot \frac{6561}{4} = \frac{39366}{4} \sin (2 \theta) = 9841.5 \sin (2 \theta) \]

Final Answer