Question

Evaluate the integral by changing to cylindrical coordinates. −₅∫⁵ ₀ ​∫√²⁵−ˣ² ₀ ​∫ ²⁵−ˣ²⁻ʸ² ​√x² +y² dzdydx

Answer 1

`\(-(125)/(3)\pi\)`Evaluating this integral involves integrating (r) with respect to (z) from (
`0\) to \(√(25 - r^2)\), \(r\)`with respect to
`\(r\) from \(0\) to \(5\),` and \(\theta\) with respect to
`\(\theta\) from \(0\) to \(2\pi\).`

Step-by-step explanation:

Converting the given triple integral into cylindrical coordinates involves substituting
`\(x = r \cos \theta\), \(y = r \sin \theta\), and \(z = z\)` and also replacing the differential volume element \(dV\) with \(r dz dr d\theta\). Integrating with respect to \(z\) from
`\(0\) to \(√(25 - x^2 - y^2)\),` with \(x\) varying from
`\(-5\) to \(5\)`and \(y\) varying from
`\(0\) to \(√(25 - x^2)\), and \(z\`) ranging from
`\(0\) to \(√(25 - x^2 - y^2)\)`within the bounds, simplifies the integral.

By converting to cylindrical coordinates, the integral becomes
`\(-\int_(0)^(2\pi) \int_(0)^(5) \int_(0)^(√(25 - r^2))`r dz dr d\theta\).

Evaluating this integral involves integrating (r) with respect to (z) from (
`0\) to \(√(25 - r^2)\), \(r\)`with respect to
`\(r\) from \(0\) to \(5\),` and \(\theta\) with respect to
`\(\theta\) from \(0\) to \(2\pi\).`The resulting computation yields
`\(-(125)/(3)\pi\), w`hich is the value of the given triple integral in cylindrical coordinates.

Answer 2

Let's start by expressing the variables x, y, and z in terms of cylindrical coordinates ρ, θ, and z: is
`\((78125)/(6) \pi\).`

`\[ x = \rho \cos(\theta) \]`

`\[ y = \rho \sin(\theta) \]`

`\[ z = z \]`

Now, let's express the given integral in terms of cylindrical coordinates:

`\[ \int_(-5)^(5) \int_(0)^(√(25-x^2)) \int_(0)^(√(25-x^2-y^2)) √(x^2 + y^2) \, dz \, dy \, dx \]`

Using the cylindrical coordinates transformations, this becomes:

`\[ \int_(0)^(2\pi) \int_(0)^(5) \int_(0)^(√(25-\rho^2)) \rho √(\rho^2+z^2) \, dz \, d\rho \, d\theta \]`

Now, we can evaluate this triple integral.

`\[ \int_(0)^(2\pi) \int_(0)^(5) \left[ (1)/(3) \rho^3 √(\rho^2+z^2) \right]_(0)^(√(25-\rho^2)) \, d\rho \, d\theta \]`

Simplifying the expression inside the brackets:

⇒
`\[ \int_(0)^(2\pi) \int_(0)^(5) \left[ (1)/(3) \left( \rho^3 √(\rho^2+25-\rho^2) - 0 \right) \right] \, d\rho \, d\theta \]`

⇒
`\[ \int_(0)^(2\pi) \int_(0)^(5) (1)/(3) \rho^3 √(25) \, d\rho \, d\theta \]`

⇒
`\[ (25)/(3) \int_(0)^(2\pi) \int_(0)^(5) \rho^3 \, d\rho \, d\theta \]`

Now, evaluate the inner integral with respect to ρ:

⇒
`\[ (25)/(3) \int_(0)^(2\pi) \left[ (1)/(4) \rho^4 \right]_(0)^(5) \, d\theta \]`

⇒
`\[ (25)/(3) \int_(0)^(2\pi) (1)/(4) (5)^4 \, d\theta \]`

⇒
`\[ (25)/(3) \int_(0)^(2\pi) (1)/(4) (625) \, d\theta \]`

⇒
`\[ (25)/(3) \int_(0)^(2\pi) (625)/(4) \, d\theta \]`

⇒
`\[ (25)/(3) \left[ (625)/(4) \theta \right]_(0)^(2\pi) \]`

⇒
`\[ (25)/(3) \cdot (625)/(4) \cdot 2\pi \]`

⇒
`\[ (25)/(3) \cdot (3125)/(2) \pi \]`

⇒
`\[ (78125)/(6) \pi \]`

So, the value of the given integral in cylindrical coordinates is
`\((78125)/(6) \pi\).`