Evaluate the integral by changing to cylindrical coordinates.

Integral²_-2 integral^squareroot 4 – y²_-squareroot 4 – y² integral²_squareroot x² + y² xz dz dx dy

integral³_-3 integral^squareroot 9 – x²₀ integral⁹-x²-y²₀ squareroot x² + y² dz dy dx

The Correct Answer and Explanation is:

To evaluate the given integrals by changing to cylindrical coordinates, we proceed as follows:

Integral 1: ∫−22∫−4−y24−y2∫x2+y24xz dz dx dy\int_{-2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \int_{x^2+y^2}^{4} xz \, dz \, dx \, dy(Brainly)

Step 1: Convert to cylindrical coordinates

In cylindrical coordinates, x=rcos⁡θx = r \cos\theta, y=rsin⁡θy = r \sin\theta, and z=zz = z. The differential element dx dydx \, dy transforms to r dr dθr \, dr \, d\theta.(Brainly)
The integrand xzxz becomes rcos⁡θ⋅zr \cos\theta \cdot z.
The limits for rr range from 0 to 2 (since x2+y2≤4x^2 + y^2 \leq 4), and θ\theta ranges from 0 to 2π2\pi.

Step 2: Set up the integral

The integral becomes:

Step 3: Evaluate the integral

First, integrate with respect to zz:

Now, the integral becomes:

Next, integrate with respect to rr:

Finally, integrate with respect to θ\theta:(Mathematics Stack Exchange)

Since the integral of cos⁡θ\cos\theta over a full period is zero, the entire integral evaluates to 0.

Integral 2: ∫−33∫09−x2∫09−x2−y2x2+y2 dz dy dx\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{\sqrt{9-x^2-y^2}} \sqrt{x^2 + y^2} \, dz \, dy \, dx(Mathematics Stack Exchange)

Step 1: Convert to cylindrical coordinates

In cylindrical coordinates, x=rcos⁡θx = r \cos\theta, y=rsin⁡θy = r \sin\theta, and z=zz = z. The differential element dx dydx \, dy transforms to r dr dθr \, dr \, d\theta.(Brainly)
The integrand x2+y2\sqrt{x^2 + y^2} becomes rr.
The limits for rr range from 0 to 3 (since x2+y2≤9x^2 + y^2 \leq 9), and θ\theta ranges from 0 to 2π2\pi.

Step 2: Set up the integral

The integral becomes:

Step 3: Evaluate the integral

First, integrate with respect to zz:

Now, the integral becomes:

Next, integrate with respect to rr:

This integral requires a substitution. Let u=9−r2u = 9 – r^2, so du=−2r drdu = -2r \, dr. The limits change accordingly: when r=0r = 0, u=9u = 9; and when r=3r = 3, u=0u = 0. The integral becomes:

Finally, integrate with respect to θ\theta:

Multiplying all parts together:

Conclusion

The value of the first integral is 0.
The value of the second integral is 18π18\pi.

These results demonstrate the application of cylindrical coordinates to simplify and evaluate double and triple integrals over circular and cylindrical regions.

Integral 1: ∫−22∫−4−y24−y2∫x2+y24xz dz dx dy\int_{-2}^{2} \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} \int_{x^2+y^2}^{4} xz \, dz \, dx \, dy(Brainly)

Integral 2: ∫−33∫09−x2∫09−x2−y2x2+y2 dz dy dx\int_{-3}^{3} \int_{0}^{\sqrt{9-x^2}} \int_{0}^{\sqrt{9-x^2-y^2}} \sqrt{x^2 + y^2} \, dz \, dy \, dx(Mathematics Stack Exchange)

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