1. ## A few questions.

1. A sphere with radius 2 ($x^2+y^2+z^2=4$) has its cylindrical core of radius 1 removed ($x^2+y^2=1$), what is the volume of the resultant solid?

a) Use cylindrical coordinates to find the volume of the resultant solid.
b) Use spherical coordinates to find the volume of the resultant solid.

For a) I just used cylindrical coordinates to find the volume of the cylinder, then used $V = \frac{4}{3}\pi r^2$ which is the volume of the sphere then minus the volume of the cylinder from it.

For b) I used spherical coordinates to find the volume of the sphere, then used $V = \pi r^2 h$ which is the volume of the cylinder, then used the volume of sphere minus the volume of cylinder.

Now what I am wondering is... is that the right way to interpret the question? Or does the question mean use cylindrical coordinates ONLY to work out the volume? If so... how do I do that? Cause I thought as long as I used cylindrical coordinates to find the volume that's fine.

Same goes for b)

2. $\mathbf{F} = (x+y)i+(y+z)j+(z+x)k$. Verify the Divergence Theorem by showing $\int \int \int_E \Delta . \mathbf{F} dV = \int \int_S \mathbf{F}.\mathbf{n} dS$ where S is the union of the surfaces $z=0$ and $z = 4-x^2-y^2$ and E is the solid encompassed by those 2 surfaces.

Now I can evaluate the LHS which in spherical coordinates is given by: $\int_0^{\frac{\pi}{2}} \int_0^{2\pi} \int_0^4 3p^2\sin(\phi)dp d\theta d\phi = 128\pi$ (btw is this triple integral right? As in did I set it up right, not the answer )

Now how do I evaluate the RHS for the surface integral over the surface $z = 4-x^2-y^2$ (without a calculator)... it's almost impossible to do by hand after you compute $\mathbf{F}(\mathbf{r}(u,v)).(\mathbf{r}_u \times \mathbf{r}_v)$ where u and v are x and y respectively.

3. Evaluate $\int_1^2 \int_{\frac{1}{y}}^{\sqrt{y}} e^{\sqrt{xy}}dxdy$ by making the transformation $x = \frac{u}{v}$ and $y = uv$

I swear you can NOT do this question without a CAS.

After sketching on the u-v plane (with u as the vertical axis) we get the region to be the region between $u = v^3, u = \frac{2}{v}$ and $u = 1$ (I've checked this many times)

Now the Jacobian is $\frac{2u}{v}$ and after making the transformations we can evaluate the integral in the u-v plane with: $\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?

2. Originally Posted by usagi_killer
1. A sphere with radius 2 ($x^2+y^2+z^2=4$) has its cylindrical core of radius 1 removed ($x^2+y^2=1$), what is the volume of the resultant solid?

a) Use cylindrical coordinates to find the volume of the resultant solid.
b) Use spherical coordinates to find the volume of the resultant solid.

For a) I just used cylindrical coordinates to find the volume of the cylinder, then used $V = \frac{4}{3}\pi r^2$ which is the volume of the sphere then minus the volume of the cylinder from it.

For b) I used spherical coordinates to find the volume of the sphere, then used $V = \pi r^2 h$ which is the volume of the cylinder, then used the volume of sphere minus the volume of cylinder.

Now what I am wondering is... is that the right way to interpret the question? Or does the question mean use cylindrical coordinates ONLY to work out the volume? If so... how do I do that? Cause I thought as long as I used cylindrical coordinates to find the volume that's fine.

Same goes for b)
The thing that you need to be careful about here is that the "cylindrical core" is not really a cylinder, because its top and bottom surfaces are not flat (they form part of the sphere). So its volume is not $\pi r^2h$. Using cylindrical polars, I get the volume of the resultant solid (sphere with "cylinder" removed) to be $4\sqrt3\pi$.

Originally Posted by usagi_killer
3. Evaluate $\int_1^2 \int_{\frac{1}{y}}^{\sqrt{y}} e^{\sqrt{xy}}dxdy$ by making the transformation $x = \frac{u}{v}$ and $y = uv$

I swear you can NOT do this question without a CAS.

After sketching on the u-v plane (with u as the vertical axis) we get the region to be the region between $u = v^3, u = \frac{2}{v}$ and $u = 1$ (I've checked this many times)

Now the Jacobian is $\frac{2u}{v}$ and after making the transformations we can evaluate the integral in the u-v plane with: $\int_1^{2^{\frac{3}{4}}} \int_{u^{\frac{1}{3}}}^{\frac{2}{u}} 2ue^u dvdu$

Now it's impossible to integrate that without a calculator.

Then we can try reverse the order of the integral so we get a dudv instead of dvdu, it's also unable to be integrated by hand (try it yourself). So is there ANYWAY to do this question by hand?
Shouldn't the integral be $\displaystyle\int_1^{2^{3/4}}\!\!\!\int_{u^{1/3}}^{2/u}\tfrac{2u}ve^u\,dvdu$ (even worse than you thought, with that v in the denominator)? Assuming that you have copied the question correctly, it certainly looks impossible to integrate analytically.