Home » Space Vehicles Express » Evaluating $\Int_{-4} ^4\Int _0 ^{\Sqrt{16-X^2}} \Int _0 ^{16-X^2-Y^2} \Sqrt{X^2 + Y^2}\, Dz\, Dy\, Dx$

Evaluating $\Int_{-4} ^4\Int _0 ^{\Sqrt{16-X^2}} \Int _0 ^{16-X^2-Y^2} \Sqrt{X^2 + Y^2}\, Dz\, Dy\, Dx$

Sophia Al-Mansoor • • 2 min read
Evaluating $\Int_{-4} ^4\Int _0 ^{\Sqrt{16-X^2}} \Int _0 ^{16-X^2-Y^2} \Sqrt{X^2 + Y^2}\, Dz\, Dy\, Dx$
$\begingroup$

Question: Evaluate the given triple integral with cylindrical coordinates:

$$\int_{-4} ^4\int _0 ^{\sqrt{16-x^2}} \int _0 ^{16-x^2-y^2} \sqrt{x^2 + y^2}\,dz\,dy\,dx$$

My solution (attempt): Upon converting the triple integral into cylindrical coordinates, I got:

$$\int_{0} ^{\pi/2}\int _0 ^{4} \int _0 ^{16-r^2\cos^2\theta-r^2\sin^2\theta} r^2\,dz\,dr\,d\theta$$

After solving for the solution, I got $1024\pi/15$ as the answer. Apparently, this is incorrect. Could someone please show me where integral conversion is wrong?

Thanks in advance!

$\endgroup$

1 Answer

$\begingroup$

First integral should be from $0$ to $\pi$ because $-4 \le x \le 4$ and $0 \le y \le \sqrt{16 - x^2}$ is a semicircle. A full circle is $0$ to $2\pi$ and half of that is $0$ to $\pi$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Sophia Al-Mansoor
Author

Sophia Al-Mansoor

Sophia analyzes international trade, startup ecosystems, retail transformation, and supply chain logistics for modern digital publications.