Questions: where E is the region with x ≥ 0 and y ≥ 0 ∭ E xy dV √(x^2 + y^2) ≤ z ≤ 1

Solution Steps

To solve this problem, we need to evaluate a triple integral over a specified region \( E \). The region \( E \) is defined by \( x \geq 0 \), \( y \geq 0 \), and the bounds for \( z \) are given by \( \sqrt{x^2 + y^2} \leq z \leq 1 \). The integrand is \( xy \). We can use cylindrical coordinates to simplify the integration process, where \( x = r \cos \theta \), \( y = r \sin \theta \), and \( z = z \). The bounds for \( r \) will be from 0 to 1, \( \theta \) from 0 to \(\pi/2\), and \( z \) from \( r \) to 1.

We are tasked with evaluating the triple integral

\[ \iiint_{E} xy \, dV \]

where \( E \) is defined by \( x \geq 0 \), \( y \geq 0 \), and \( \sqrt{x^2 + y^2} \leq z \leq 1 \). In cylindrical coordinates, we have \( x = r \cos \theta \), \( y = r \sin \theta \), and the volume element \( dV = r \, dr \, d\theta \, dz \). Thus, the integrand becomes

\[ xy = r^2 \cos \theta \sin \theta. \]

The limits for the integration are as follows:

• For \( r \): from \( 0 \) to \( 1 \)
• For \( \theta \): from \( 0 \) to \( \frac{\pi}{2} \)
• For \( z \): from \( r \) to \( 1 \)

We first integrate with respect to \( z \):

\[ \int_{r}^{1} r^3 \sin \theta \cos \theta \, dz = r^3 \sin \theta \cos \theta (1 - r). \]

Next, we integrate with respect to \( r \):

\[ \int_{0}^{1} r^3 (1 - r) \, dr = \frac{1}{40}. \]

Finally, we integrate with respect to \( \theta \):

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{40} \sin \theta \cos \theta \, d\theta = \frac{1}{40} \cdot \frac{1}{2} = \frac{1}{80}. \]

Final Answer