[Math] Finding the volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$

calculusintegrationlimitsmultivariable-calculusvolume

The problem is stated below:

Let $V$ be volume bounded by surface $y^2=4ax$ and the planes $x+z=a$ and $z=0$.

Express $V$ as a multiple integral, whose limits should be clearly stated. Hence calculate $V$.

Progress

I want to find out the limits of the multiple integral that's needed to calculate $V$.

I'm guessing that:
$x=a-z$, $x=(y^2)/4a$

$$y= \pm \sqrt{4ax},\quad z=0,\quad z=a-x $$

but it seems like I've used the upper plane equation twice?

Also, it would really help if we could compare our answer volumes to check that this is right from the start!

Thanks everyone!!! 🙂

Best Answer

Sometimes a picture is worth a thousand words. Eventually, a math student should try plot graphs in their minds with out the use of software (although, I've always preferred Play Doh because after solving the problem I could always make little fishies and green froggies).

Geometrically, Your problem should look like this... enter image description here

We see that $x=a$ when the upper plane intersect the $x-y$ plane. We also see that when $x=t$ $$y^2=4at$$ $$y=\pm2\sqrt{at}$$

So the integral becomes $$\begin{array}{lll} \int^a_0\int_{-2\sqrt{ax}}^{2\sqrt{ax}}(a-x)dydx&=&\int^a_0(a-x)(4\sqrt{ax})dx\\ &=&\int^a_0(4a^\frac{3}{2}x^\frac{1}{2} - 4a^\frac{1}{2}x^\frac{3}{2})dx\\ &=&(\frac{2}{3}\cdot 4a^\frac{3}{2}a^\frac{3}{2} - \frac{2}{5}\cdot 4a^\frac{1}{2}a^\frac{5}{2})\\ &=&8a^3(\frac{1}{3}-\frac{1}{5})\\ &=&8a^3(\frac{2}{15})\\ &=&\frac{16}{15}a^3\\ \end{array}$$