Q8. Evaluate ∭E z dV. E is the solid bounded by z=2-x^2-y^2 and z=1.

To evaluate the triple integral \(\iiint_E z \, dV\), where \(E\) is the solid bounded by \(z = 2 - x^2 - y^2\) and \(z = 1\), we will use cylindrical coordinates for simplicity.

In cylindrical coordinates \((r, \theta, z)\), the boundaries transform as follows:
- \(x = r \cos \theta\)
- \(y = r \sin \theta\)
- \(z = z\)
- The Jacobian determinant for the transformation to cylindrical coordinates is \(r\).

The solid \(E\) is bounded by \(z = 1\) (bottom surface) and \(z = 2 - r^2\) (top surface).

1. The bounds for \(z\) are from \(1\) to \(2 - r^2\).
2. The bounds for \(r\) are determined by the projection of the region onto the \(xy\)-plane. At \(z = 1\), \(1 = 2 - r^2\), so \(r = \sqrt{1} = 1\). Thus, \(r\) ranges from \(0\) to \(1\).
3. The bounds for \(\theta\) are from \(0\) to \(2\pi\).

The integral in cylindrical coordinates becomes:
\[
\iiint_E z \, dV = \int_0^{2\pi} \int_0^1 \int_1^{2-r^2} z \, r \, dz \, dr \, d\theta
\]

First, integrate with respect to \(z\):
\[
\int_1^{2-r^2} z \, dz = \left[ \frac{z^2}{2} \right]_1^{2-r^2} = \frac{(2 - r^2)^2}{2} - \frac{1^2}{2}
= \frac{(4 - 4r^2 + r^4)}{2} - \frac{1}{2}
= 2 - 2r^2 + \frac{r^4}{2} - \frac{1}{2}
= \frac{3}{2} - 2r^2 + \frac{r^4}{2}
\]

Now, the integral becomes:
\[
\int_0^{2\pi} \int_0^1 \left( \frac{3}{2} - 2r^2 + \frac{r^4}{2} \right) r \, dr \, d\theta
\]

Separate the integral:
\[
\int_0^{2\pi} d\theta \int_0^1 \left( \frac{3}{2}r - 2r^3 + \frac{r^5}{2} \right) dr
\]

Evaluate the \(\theta\) integral:
\[
\int_0^{2\pi} d\theta = 2\pi
\]

Now, evaluate the \(r\) integral:
\[
\int_0^1 \left( \frac{3}{2}r - 2r^3 + \frac{r^5

Learn how to evaluate the triple integral ∭E z dV, where E is bounded by z=2-x^2-y^2 and z=1. This problem involves applying cylindrical coordinates and integrating over a specific 3D region. Suitable for advanced undergraduate mathematics students.

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