Volume between a region (above x-axis and below parabola) and a surface

[songoku]

Homework Statement

Let R be the part of the xy-plane above the x-axis and below the parabola $y=1-x^2$. Find
the volume between R and the surface $z=x^2 \sqrt{1-y}$

Relevant Equations

Double Integral

Integration in the order of dy then dx:
$$\int_{-1}^{1} \int_{0}^{1-x^2} x^2 \sqrt{1-y} ~dy~dx$$
$$=\int_{-1}^{1} -x^2 \left[\frac{2}{3} (1-y)^{\frac{3}{2}}\right]_{0}^{1-x^2}dx$$
$$=\int_{-1}^{1}\left(-\frac{2}{3}x^5 + \frac{2}{3} x^2\right)dx$$
$$=\left. -\frac{1}{9} x^6 + \frac{2}{9} x^3\right|_{-1}^{1}$$
$$=\frac{4}{9}$$

Integration in the order of dx then dy:
$$\int_{0}^{1} \int_{-\sqrt{1-y}}^{\sqrt{1-y}}x^2\sqrt{1-y}~dx~dy$$
$$=2\int_{0}^{1} \int_{0}^{\sqrt{1-y}}x^2\sqrt{1-y}~dx~dy$$
$$=2\int_{0}^{1} \sqrt{1-y} \left[\frac{1}{3} x^3\right]_{0}^{\sqrt{1-y}}dy$$
$$=\frac{2}{3}\int_{0}^{1}(1-y)^2 dy$$
$$=-\frac{2}{9} \left. (1-y)^3\right|_{0}^{1}$$
$$=\frac{2}{9}$$

I got two different results. Where is my mistake?

Thanks

 

[nuuskur]

Both integrals are correctly parametrised. The result of the second integral is correct.

In the first integral you should evaluate instead to
<br /> \sqrt{(1-y)^3} \Bigg\vert _0^{1-x^2} = \sqrt{x^6} - 1 = |x|^3 -1<br />
yielding
<br /> -\frac{2}{3}\int _{-1}^1 x^2(|x|^3-1)dx = \frac{2}{9},<br />
as required.

 

[songoku]

In the first integral you should evaluate instead to
<br /> \sqrt{(1-y)^3} \Bigg\vert _0^{1-x^2} = \sqrt{x^6} - 1 = |x|^3 -1<br />
yielding
<br /> -\frac{2}{3}\int _{-1}^1 x^2(|x|^3-1)dx = \frac{2}{9},<br />
as required.

How to know that the correct working should be like this, not like the one I did?

Thanks

 

[nuuskur]

Because it's always been the case that $\sqrt{a^2} = |a|$.

Visually, if you take the integral first w.r.t $y$, you are finding the "area" of an "infinitely thin" slice of the body of interest along the $x$-axis. But if you don't take absolute value, you lose symmetry w.r.t the plane $x=0$. In other words,
<br /> -\frac{2}{3} \int_{-1}^0 x^2(x^3-1)dx \neq -\frac{2}{3} \int_{0}^1 x^2(x^3-1)dx.<br />

 

[songoku]

Thank you very much for the help and explanation nuuskur