Volume between surfaces 
Find $V(T),$ where $T$ is the region bounded by the surfaces $y = kx^2+kz^2$ and $z=kx^2+ky^2,$ where $k\in\mathbb{R}, k > 0.$

I tried solving for the area over which these curves intersect, which gave me $y-kz^2 = z-ky^2.$ Solving gives $y+ky^2 - (z+kz^2) =0\Rightarrow (y-z)(1 + k(y-z)) = 0.$ Since $y$ and $z$ are nonnegative, this implies $y=z.$ We thus obtain $y = kx^2 + ky^2\Rightarrow ky^2 - y +kx^2 = 0\Rightarrow (y -\dfrac{1}{2k})^2 +x^2 = \dfrac{1}{4k^2},$ which is a circle centered at $(0,\dfrac{1}{2k})$ with radius $\dfrac{1}{2k}.$ 
I am able to solve for $A(x),$ the area in terms of $x$ at a given value of $x,$ and I know $x$ ranges from $-\dfrac{1}{2k}$ to $\dfrac{1}{2k},$ but the resulting integral I have to evaluate is absolutely disgusting!
Is there a "cleaner" integral I can use?

Observe that the enclosed volume by $y = kx^2+kz^2$ and $z=kx^2+ky^2$ are symmetric with respect to the plane $y=z$. So, the total volume is twice the volume between the surfaces $y=z$ and $z=kx^2+ky^2$. 
Recognize that the integration region in $xy$-coordinates is the circle given by 
$$x^2+ \left(y -\frac{1}{2k}\right)^2 = \dfrac{1}{4k^2}$$
and re-center the circle with the variable changes $x=u$ and $v = y - \frac{1}{2k}$. Then, the integration region becomes $u^2+ v^2 = \frac{1}{4k^2}$ and the two enclosing surfaces are respectively,
$$z_1 = v+ \frac 1{2k},\>\>\>\>\> z_2=ku^2+k\left(v+\frac1{2k}\right)^2$$
As a result, the volume integral can be expressed as,
$$V= 2\int_{u^2+v^2\le \frac1{4k^2}} (z_1-z_2)dudv
= 2\int_{u^2+v^2\le \frac1{4k^2}} (\frac1{4k}-ku^2-kv^2)dudv$$
Then, integrate in polar coordinates to obtain,
$$V=2\int_0^{2\pi} \int_0^{\frac1{2k}}(\frac1{4k}-kr^2)rdrd\theta
=4\pi\int_0^{\frac1{2k}}(\frac1{4k}-kr^2)rdr =\frac\pi{16k^3} $$
