How is the triple integral of $(x^2+y^2)$ over the solid region bounded by $z=x^2+y^2$ and $z=4$ evaluated using the slicing method?

Q: How is the triple integral of $(x^2+y^2)$ over the solid region bounded by $z=x^2+y^2$ and $z=4$ evaluated using the slicing method?

The integral is evaluated by integrating with respect to $x$ and $y$ first over a horizontal disk cross-section, and then integrating with respect to $z$. Polar coordinates are used for the inner double integral, yielding an integrand of $r^3$. The final result is $\frac{32\pi}{3}$.

The integral is evaluated by integrating with respect to

x

and

y

first over a horizontal disk cross-section, and then integrating with respect to

z

. Polar coordinates are used for the inner double integral, yielding an integrand of

r^3

. The final result is

\frac{32\pi}{3}

Conditions

The solid region $\Omega$ is bounded below by the paraboloid $z=x^2+y^2$ and above by the plane $z=4$ .
The integrand is $f(x,y,z) = x^2+y^2$ .
The integration order is $dx\,dy\,dz$ (slicing method).

Reasoning, step by step

Identify the bounds for the outer integral: $z$ ranges from $0$ to $4$ .
For a fixed $z$ , the cross-section $D(z)$ is a disk defined by $x^2+y^2 \le z$ .
Convert the inner double integral over $D(z)$ to polar coordinates: $x = r\cos\theta$ , $y = r\sin\theta$ .
Substitute $x^2+y^2 = r^2$ and the area element $dx\,dy = r\,dr\,d\theta$ .
Set the polar limits: $\theta$ ranges from $0$ to $2\pi$ , and $r$ ranges from $0$ to $\sqrt{z}$ .
Evaluate the inner integral: $\int_0^{2\pi} d\theta \int_0^{\sqrt{z}} r^2 \cdot r \, dr = 2\pi \left[ \frac{r^4}{4} \right]_0^{\sqrt{z}} = \frac{\pi}{2} z^2$ .
Evaluate the outer integral: $\int_0^4 \frac{\pi}{2} z^2 \, dz = \frac{\pi}{2} \left[ \frac{z^3}{3} \right]_0^4 = \frac{32\pi}{3}$ .

Example

The script states: 'The outer integral runs from $z=0$ to $z=4$ . For the inner double integral over the disk $D(z)$ , polar coordinates are used where the angle goes from $0$ to $2\pi$ and the radius $r$ goes from $0$ to $\sqrt{z}$ . Substituting $x^2+y^2=r^2$ and the Jacobian $r$ , the integrand becomes $r^3$ . Evaluating this gives $\frac{1}{4}z^2$ , which leads to the final answer of $\frac{32\pi}{3}$ after integrating along $z$ .','

Common misconceptions

Assuming the radius of the cross-section is constant; it depends on $z$ as $r=\sqrt{z}$ .
Forgetting the Jacobian factor $r$ when converting to polar coordinates.
Integrating with respect to $z$ first, which would require a different setup (cylindrical shells or projecting onto the xy-plane).

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How is the triple integral of $(x^2+y^2)$ over the solid region bounded by $z=x^2+y^2$ and $z=4$ evaluated using the slicing method?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

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Related questions

How is the triple integral of (x2+y2)(x^2+y^2)(x2+y2) over the solid region bounded by z=x2+y2z=x^2+y^2z=x2+y2 and z=4z=4z=4 evaluated using the slicing method?

Conditions

Reasoning, step by step

Example

Common misconceptions

Watch the explanation

Connected concepts

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How is the triple integral of $(x^2+y^2)$ over the solid region bounded by $z=x^2+y^2$ and $z=4$ evaluated using the slicing method?