What is the volume of the solid in $xyz$-space?

Question

What is the volume of the solid in $xyz$-space bounded by the surfaces $y = x^2$, $y=2$, $z=0$, and $z=y+3$ ?

Answer 1

I think the easier is to work at $y$ constant (i.e. final integration in $y$).

The nominal section is a rectangle $L\times H$ if I represent it correctly in my mind with

$L=2x=2\sqrt y$

$H=z=y+3$

So I think $V=\int_0^2 LH\;dy=\int_0^2 2\sqrt y(y+3)\;dy=\bigg[\frac 45y^2\sqrt y+4y\sqrt y\bigg]_0^2=\sqrt 2(\frac{16}{5}+8)=\frac{56}{5}\sqrt 2$

We could have written it $\displaystyle V={\int_0^2}\int_0^{y+3}\int_{-\sqrt y}^{\sqrt y} dx\;dz\;dy\qquad$ (with the integration done in the order of the little $dv_{ar}$).

Remark :

If you substitute all in $z$ as you suggested to get $z=x^2+3$ and work at $x$ constant, the section is now a trapezoid, so it is a little less easier to integrate.

$\displaystyle V=\int_{-\sqrt 2}^{\sqrt 2}\int_{x^2}^{2}\int_0^{y+3}dz\;dy\;dx\qquad$

The worse would be to work at $z$ constant, in this case the section is a flattened/truncated parabola, quite ugly to integrate, I do not even want to write the formula it is a sum of many integrals.

But in theory, all $3$ methods should lead to the same result, select the one that seems the easiest.

As an exercise try to calculate the second formula for $V$ and see that you get the same value.

Answer 2

The body in question is cylinder which its section base in plane XY is the region of a parabola y=x^2 limited by the plane y=2. The best approach in my opinion is cutting our body by planes parallel to plane XZ and then consider the rectangular section resulting. The area of this rectangular section is 2x*(y+3). By substitution of x for square root of y. You get the simple integral of dV= 2(y+3)*y^(1/2)dy limit 0 and 2. I leave the work here.