I Which of the two answers is correct?

[WMDhamnekar]

TL;DR

Finding the volume above cone $z= a\sqrt{x^2 + y^2}$ and inside the sphere $x^2 + y^2 + z^2 = b^2$

Solution 1:

The answer is $\frac{2b^5\pi}{5} \times \left(1 -\frac{a}{\sqrt{1+a^2}}\right)$

Solution 2:

I want to decide which answer is correct? Would you help me in this task?

 

[Delta2]

I am with solution 2, easier to understand for me and I find it correct

 

[Delta2]

After a closer look at Sol1. it computes not the volume V but the integral $$\iiint_V (x^2+y^2+z^2)dxdydz$$. The result of Sol1. would be the same as Sol2. if instead it would calculate $$\iiint_V 1 dxdydz=\int_0^{2\pi}\int_0^a\int_0^b \rho^2\sin\phi d\rho d\phi d\theta$$

 

[WMDhamnekar]

After a closer look at Sol1. it computes not the volume V but the integral $$\iiint_V (x^2+y^2+z^2)dxdydz$$. The result of Sol1. would be the same as Sol2. if instead it would calculate $$\iiint_V 1 dxdydz=\int_0^{2\pi}\int_0^a\int_0^b \rho^2\sin\phi d\rho d\phi d\theta$$

Thanks for your scrutiny of both the solutions. But I want more confirmation from other elite members.

 

[Delta2]

Thanks for your scrutiny of both the solutions. But I want more confirmation from other elite members.

Well ok, just to emphasize that the two solutions don't compute the same thing

We have a region V (as the region between the surface of a cone and a sphere).

Sol 1 computes $\iiint_V (x^2+y^2+z^2)dxdydz$
Sol 2 computes $\iiint_V 1 dxdydz$

 

[hutchphd]

Yes. Why do the solutions have different units ? Junk.

 

[Delta2]

Yes. Why do the solutions have different units ? Junk.

Check post #5.

 

[hutchphd]

Yes.