Evaluate the integral by changing to cylindrical coordinates: $\int_{-3}^3 \int_{-\sqrt{9 - y^2}}^{\sqrt{9 - y^2}} \int_0^6 xz \, dz \, dx \, dy$

Neetesh Kumar | November 26, 2024

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This is the solution to Math 1D
Assignment: 15.7 Question Number 15
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Step-by-step solution:

Step 1: Understanding the given integral

The given integral is a triple integral: $\int_{-3}^3 \int_{-\sqrt{9 - y^2}}^{\sqrt{9 - y^2}} \int_0^6 xz \, dz \, dx \, dy$

The limits of integration indicate:

• For $y$, the range is $-3 \leq y \leq 3$.
• For $x$, the range depends on $y$: $-\sqrt{9 - y^2} \leq x \leq \sqrt{9 - y^2}$, representing a circle in the $x$-$y$ plane with radius $3$.
• For $z$, the range is $0 \leq z \leq 6$.

Step 2: Convert to cylindrical coordinates

Cylindrical coordinates are given by:

• $x = r \cos\theta$, $y = r \sin\theta$, and $x^2 + y^2 = r^2$.
• The volume element is $dV = r \, dr \, d\theta \, dz$.

In cylindrical coordinates:

• The circle $x^2 + y^2 = 9$ becomes $r^2 = 9$, so $0 \leq r \leq 3$.
• $z$ remains $0 \leq z \leq 6$.
• The angle $\theta$ spans the full circle: $0 \leq \theta \leq 2\pi$.

The integrand $xz$ becomes: $xz = (r\cos\theta)z$

Step 3: Rewrite the integral

The integral in cylindrical coordinates becomes: $\int_0^{2\pi} \int_0^3 \int_0^6 (r\cos\theta)z \cdot r \, dz \, dr \, d\theta$

Simplify the integrand: $r \cdot (r\cos\theta)z = r^2 \cos\theta z$

Thus, the integral is: $\int_0^{2\pi} \int_0^3 \int_0^6 r^2 \cos\theta z \, dz \, dr \, d\theta$

Step 4: Evaluate the integral step-by-step

1. Innermost integral (over $z$): $\int_0^6 z \, dz = \left[\frac{z^2}{2}\right]_0^6 = \frac{6^2}{2} - \frac{0^2}{2} = 18$

Substitute back: $\int_0^{2\pi} \int_0^3 r^2 \cos\theta \cdot 18 \, dr \, d\theta$

2. Factor out constants: $18 \int_0^{2\pi} \int_0^3 r^2 \cos\theta \, dr \, d\theta$

3. Second integral (over $r$): $\int_0^3 r^2 \, dr = \left[\frac{r^3}{3}\right]_0^3 = \frac{3^3}{3} - \frac{0^3}{3} = 9$

Substitute back: $18 \cdot 9 \int_0^{2\pi} \cos\theta \, d\theta$

4. Outer integral (over $\theta$): The integral of $\cos\theta$ over a full period is $0$: $\int_0^{2\pi} \cos\theta \, d\theta = 0$

Thus, the entire integral evaluates to: $0$

Final Answer:

The value of the given integral is: $\boxed{0}$

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