How to get the volume of a solid by triple integration , which is formed by intersection of two surfaces? Namely $z = x^2 + y^2$ and $z = 4 y$.

Question

How to get the volume of a solid by triple integration , which is formed by intersection of two surfaces? Namely $z = x^2 + y^2$ and $z = 4 y$.

I am confused with limits, $z = x^2 + y^2$ and $z = 4y$.

Answer 1

Let $y=u+2$, so the equations become $$z=x^2+u^2+4u+4$$ $$z=4u+8$$ So every point of the intersection holds: $$x^2+u^2=4$$ Then, the volume is $$\int_{x^2+u^2\le4}[4u+8-(x^2+u^2+4u+4)]dxdu$$ $$=\int_{x^2+u^2\le 4}(4-x^2-u^2)dxdu$$ $$=\int_0^{2\pi}\int_0^2\rho(4-\rho^2)d\rho d\theta$$ $$=2\pi\left(2\cdot2^2-\frac{2^4}4\right)=8\pi$$

Answer 2

We need at first some sketch for the domain, I suggest to consider the $z-x$ plane for example.

Then by intersection we find

• $x^2+y^2=4y \implies x^2+(y-2)^2=4$

therefore on the $x-y$ plane the domain is a circle with radius $2$ centered at $(0,2)$.

From here we can obtain the set up by

$$\int_0^4 dy\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx \int_{x^2+y^2}^{4y}dz$$