Find the volume of the solid in $xyz$-plane bounded by $y=x^2,y=2-x^2,z=0$ and $z=y+3$.
I have found the answer $\frac {13 \pi} {6}$. Is it correct at all? Please verify it.
Thank you very much.
I have drawn it in my mind. Can anybody help me plotting it online? How do I draw it to make sure that I have judged the correct surface intuitionally?
EDIT $:$ I have found it in the following way $:$
The required volume is
$$\int_{-1}^{1} \int_{x^2} ^{2-x^2} \int_{0}^{3+y}\ \mathrm {dz\ dy\ dx}$$ i.e. the volume of the required solid is $$16 \int_{0}^{1} (1-x^2)\ \mathrm {dx}$$ which simplifies to $\frac {32} {3}$.
Now please verify my solution now.
I would set up the integral like this:
$\int_{-1}^{1}\int_{x^2}^{2-x^2}\int_0^{y+3} \ dz\ dy\ dx$
As for the figure, you have a cylinder with a cross section of two parabola. Cut at one end at an angle.
The area of the cross section is:
$\int_{-1}^{1} 2 - 2x^2 = \frac 83$
Does slicing at an angle make a difference? The cross section is symmetric about the line $y = 1.$ As much is cut off below the this line as above it.
$\int_0^4\int_{-1}^{1} 2 - 2x^2 \ dx \ dz = \frac {32}{3}$