Calculate the volume bounded by the surfaces

Question

Calculate the volume of the solid bounded by the surfaces $$\begin{aligned}z&=4x^2+4y^2, \\ z&=x^2+y^2, \\z&=4.\end{aligned}$$

I made an equation of $4x^2+4y^2=4-x^2+y^2$ and solved it to get $x^2+y^2=\dfrac{4}{5}$.

Then I did a double integration $$\displaystyle \iint_{x^2+y^2\leq \frac{4}{5}} \left(4x^2+4y^2\right)-\left(4-x^2-y^2\right)dA,\ $$ did the subtraction and the changed to polar $$\displaystyle \int_0^{2\pi} \int_0^{\frac{4}{5}} \left(3r^2-4\right) r \,dr \,d\theta,\ $$ got a result of $$ \displaystyle\int_0^{ 2\pi}\Bigg[ \left.\left(\frac{3}{4}r^4-2r^2\right) \right\vert_{r=0}^{\frac{4}{5}} \Bigg]d\theta $$

Obviously this is not correct. Can you tell me where I have gone wrong?

Thanks in advance :)

Answer 1

From your second and third equations it is clear that $0\le z\le 4$. For a given $z$, the first two equations give circles centered at $x=y=0$. So the solid you want is the region between those two circles for all values of $z$ between $0$ and $4$. (This is between two paraboloids of revolution, below the plane $z=4$.)

So for each $z$, find the area between the circles $x^2+y^2=z$ and $x^2+y^2=\frac z4$. This uses simple algebra and geometry, and you now have a simple formula for area in terms of $z$. Integrate this formula from $z=0$ to $z=4$, and you have the desired volume.

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The answer to your question "where I have gone wrong" is that you tried to get an equation relating $x$ and $y$ without using $z$. This is not possible, though you can get the equation of one of the many circles involved with the solid. But you did not know which circle that was, so everything after that was pointless.

My solution also finds two families of circles, but it knows the significance of those circles and uses them properly.

Answer 2

Let $(x,y,z) \in \mathbb{R}^{3}$ and let $S$ be the region enclosed by the surfaces $z = x^{2}+y^{2}$ and $z=4$. Then $(x,y,z) \in S$ if and only if $|x| \leq 4, |y| \leq \sqrt{4-x^{2}}, 0 \leq z \leq 4$. Thus $$4\int_{0}^{4}\int_{0}^{\sqrt{4-x^{2}}} x^{2} + y^{2} dy dx = 4\int_{0}^{4} x^{2}\sqrt{4-x^{2}} + \frac{(4-x^{2})^{3/2}}{3} dx$$ is the content of $S$.

Let $T$ be the region enclosed by $z=4$ and $z = 4x^{2} + 4y^{2}$ and do the above for $T$. Then the absolute value of the difference of the resulting two contents is the desired content.

Answer 3

Three volumes.

Say $r^2 = x^2 + y^2$, cut through the volume gives the following image.

Cylinder #1

$$ V_{C_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^4 dz r = 16 \pi $$

Paraboloid #1

$$ V_{P_1} = \int_0^{2\pi} d\phi \int_0^2 d r \int_0^{r^2} dz r = 8 \pi $$

Cylinder #2

$$ V_{C_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^4 dz r = 4 \pi $$

Paraboloid #2

$$ V_{P_1} = \int_0^{2\pi} d\phi \int_0^1 d r \int_0^{4 r^2} dz r = 2 \pi $$

End volume is

$$ \Big( V_{C_1} - V_{P_1} \Big) - \Big( V_{C_2} - V_{P_2} \Big) = 6 \pi. $$