1
$\begingroup$

I am extremely confused on this question and I am wondering if anyone can help me.

The question says:

Consider the Integral:

$$\int_0^8 \int_{-\sqrt{64-y^2}}^{\sqrt{64-y^2}}\int_{\sqrt{x^2+y^2}}^{ 8} 1\, dx\, dy \, dz\, .$$

enter image description here

I am completely lost on how to solve this problem. Any help would really be appreciated.

$\endgroup$

3 Answers 3

Reset to default
1
$\begingroup$

As we see in $$\int_0^8 \int_{-\sqrt{64-y^2}}^{\sqrt{64-y^2}}\int_{\sqrt{x^2+y^2}}^{ 8} 1\, dx\, dy \, dz$$ we have $$\sqrt{x^2+y^2}\leq z\leq8$$ $$-\sqrt{64-y^2}\leq x\leq \sqrt{64-y^2}$$ $$0\leq y\leq 8$$ clearly, this shape is a cone (incomplete) and with substituting spherical coordinates: $$0\leq \theta\leq \pi$$ $$0\leq \phi\leq \dfrac{\pi}{4}$$ $$0\leq \rho\leq \dfrac{8}{\cos\phi}$$ so $${\bf V}=\int_0^{\pi}\int_0^\frac{\pi}{4}\int_0^\frac{8}{\cos\phi}{\rho}^2\sin\phi\,d{\rho}\,d{\phi}\,d{\theta}=\color{blue}{\dfrac{256\pi}{3}}$$

$\endgroup$
1
$\begingroup$

Here's answers...

Consider the integral given. It runs on 0 to 8 for the outermost bound, then the next bound runs $\pm\sqrt{64-y^2}.$ That indicates the semicircle portion of the origin-centered circle of radius 8 that has positive y-coordinate. And the last part give the semicircle a 3-d shape.

So what is this?

It's a quarter of a sphere. Because the final limit goes to 8, the quarter-circle is definitely bounded by z=8 and x=0.

So your limits are 0 to 8 (for the radius), 0 to $\frac{\pi}{2},$ (for $\theta$), and 0 to $\pi$ (for $\phi$). Or at least I'm fairly sure.. I could have screwed up. Again.

Part (c)'s answer is obvious: We can already calculate the volume of this solid with formulas we know.

That means that the integral's result (and the volume of the solid given by the triple integral) is $\frac{\frac{4}{3}\pi(8)^3}{4}=\boxed{\frac{512\pi}{3}}.$

$\endgroup$
4
  • $\begingroup$ That is not the correct answer for the result of the integral $\endgroup$
    – Bailey
    Mar 9, 2017 at 3:29
  • $\begingroup$ so the rest of it is right? $\endgroup$
    – y_prime
    Mar 9, 2017 at 3:31
  • $\begingroup$ No none of upper limits are correct $\endgroup$
    – Bailey
    Mar 9, 2017 at 3:32
  • $\begingroup$ ok thats more useful.... oh wait i got it.. it should be a quarter of the sphere, let me change it, you can try it if you want but i might be wrong again :\ $\endgroup$
    – y_prime
    Mar 9, 2017 at 3:35
1
$\begingroup$

You have to convert the equations of the surfaces to spherical coordinates first. So $z=\sqrt{x^2+y^2}$ becomes $\phi=\frac{\pi}{4}$ (the side surface of the cone) and $\rho=0$ (the point at the origin.) $z=8$ becomes $\rho=8\sec{\phi}$

The region in the $xy$ plane appears to be the half disk where $y>0$, so your $\theta$ limits should be from $0$ to $\pi$.

The limits for $\phi$ should be from $0$ (the positive z direction) to $\frac{\pi}{4}$ (the side of the cone.)

The limits for $\rho$ should be from the inside surface $\rho=0$ to the outside surface $\rho=8\sec{\phi}$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .