# Thread: Surface area of a disc

1. ## Surface area of a disc

I have a surface defined as $z(x^2+y^2)=4$. I have calculated the surface integral but now I just want to calculate the contribution from the disc z=4. How do i do this?

2. ## Re: Surface area of a disc

Pretty straight forward, isn't it? If z= 4, your equation becomes $4(x^2+ y^2)= 4$ so $x^2+ y^2= 1$. That's a circle with radius 1.

3. ## Re: Surface area of a disc

I thought so but the answer is 15pi, not pi

5. ## Re: Surface area of a disc

The surface integral was 33pi. In part (b), I had to calculate volume integral where V is the region defined by $0 and $1. This was 48pi.

Now I have to verify divegence theorem. Well, I need to show 33pi + contribution from z=4 + contribution from z=1 = 48pi.

Sorry if I was not clear enough.

6. ## Re: Surface area of a disc

In cylindrical coordinates, $x^2+ y^2= \frac{4}{z}$ becomes $z= \frac{4}{r^2}$. Of course, in order to have a closed figure, we must have the "top" and "bottom", z= 4, and z= 1.

The divergence theorem says that $\int\oint\vec{F}d\vec{S}= \int\int\int \nabla\cdot\vec{F} dV$.

If you want to find the volume of the figure, $\int\int\int dV$, using that, you must have a vector function, $\vec{F}$, such that $\nabla\cdot\vec{F}= 1$. What vector function did you use?

Of course, the top of the figure, z= 4, is the disk $x^2+y^2\le 1$ while the bottom of the figure, z= 1, is the disk $x^2+ y^2= 4$ which means that, to find the volume directly, you must integrate the constant 1 from z= 0 to z= 4 over the disk $x^2+ y^2= 1$ which is simply 1 times 4 times the area of that disk, $4\pi$. Then, integrate the constant 1 from z= 0 to $z= \frac{4}{x^2+ y^2}= \frac{4}{r^2}$ (in cylindrical coordinates) for r= 1 to 2 (and $\theta$ from 0 to $2\pi$, of course).

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# how to caluclate area of a disc

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