Solving a 3D Integral: Finding the Volume Enclosed by x=y, z=0, and y^2+z^2=x

[nhrock3]

find the volume enclosed by x=y z=0 and $$y^2+z^2=x$$
i built

$$\int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}dzdydx$$


i tried to traw the projection of the top interval in the z-y plane

$$z=\sqrt{x-y^{2}}$$ -> $$z^2+y^2=x$$
so we have a circle
but the radius changes from x to $$\sqrt{x}$$.

which is not possible because when we scan the ring it goes from the biiger radius to the smaller one.
then i thought to try in anyway ,and i wrote the integral in polar
$$\int_{0}^{1}\int_{0}^{2\pi}\int_{x}^{\sqrt{x}}rdrd\theta dz$$

but i don't know how mathmatickly to change the intervals?
how to solve it in a polar way?

 

[SammyS]

In the x-y plane, you are bounded by the half parabola, $y=-\sqrt{x}$, for y < 0, and by y = x, for y ≥ 0 .

So, the limits on the middle integral should be: $\dots\int_{-\sqrt{x}}^{x}\ \dots\ dy\,\dots$

Added in edit:

Sorry for the above.

It looks like you are right.

You have: $\displaystyle \int_{0}^{1}\int_{x}^{\sqrt{x}}\int_{0}^{\sqrt{x-y^{2}}}(1)\,dz\,dy\,dx$

 

[SammyS]

...
i tried to draw the projection of the top interval in the z-y plane

$z=\sqrt{x-y^{2}}$ -> $z^2+y^2=x$
so we have a circle
but the radius changes from x to $\sqrt{x}$.

which is not possible because when we scan the ring it goes from the bigger radius to the smaller one...

If 0 < x < 1, then x < √(x) , ... if that's what's worrying you.

 

[nhrock3]

how to solve it?

 

[vela]

Try using an integral where you integrate with respect to x first, z second, and y last, i.e.
$$\int \int \int (\cdots)\,dx\,dz\,dy$$
If you haven't already, plot or sketch the surfaces. Setting up the integral is straightforward once you see how the surfaces intersect.

 

[nhrock3]

why?

i already build an integral
how to solve it?

 

[vela]

Then just integrate it. They're elementary integrals.

 

[nhrock3]

have you tried?
its not possible
how to solve it in polar

 

[vela]

You're right. Your integral makes a mess. If you use my suggestion, you get integrals that are easy to evaluate.

 

[Skins]

I didn't try to integrate this but a couple suggestions that sometimes work for when you get stuck with messy hard to solve integrals...

1) Try to set it up in terms of using a different order of integration...

2) Change of variables or coordinate system. Try using polar coordinates as suggested.

 

[nhrock3]

but i need to do addjustments to the intervals
if i change the order of integration
and its a 3d

could you solve it?

as you see my original question is about polar representation..
could you solve it this way?

 

[nhrock3]

it worked
how to solve
$$\int (y-y^2)^{\frac{3}{2}}$$

 

[vela]

Complete the square to write the integrand in the form $[a^2-(y-b)^2]^{3/2}$. Then use the appropriate substitutions.

 

[nhrock3]

$$\int (y-y^2)^{\frac{3}{2}}$$=$$\int [-(y-\frac{1}{2})^2+\frac{1}{4}]^{\frac{3}{2}}$$
what now?

 

[nhrock3]

solved it thanks