3ddefinite integralsintegrationmultivariable-calculus
Find the volume of the solid bounded above by the Parabolic cylinder $z=1-y^2$ and below the plane $2x+3y+z+10=0$ and on the sides of circular cylinder $x^2+y^2-x=0$
Since this is region obtained by three surfaces, i am unable to find out the limits of integration. My question is, in $3D$ plotter i observed that the parabolic cylinder and the circular cylinder are meeting.
But how to solve the equations $x^2+y^2-x=0$ and $z=1-y^2$
The cross sections perpendicular to the $x$-axis start at $x=0$ and end at $x=1$. Let $A(x)$ denote the area of the cross section perpendicular to the $x$-axis that intersects the $x$-axis at $x$. The solid can be thought of as the "sum of the cross sections" as they range from $x=0$ to $x=1$. The volume of the solid is thus $$\int_{0}^1 A(x)\,dx.$$
We need to find an explicit expression for $A(x)$ in terms of $x$. We know that the cross-section at $x$ is a square. The bottom vertices of this square are bounded by the parabola $x=1-y^2$. So, the side length $\color{maroon}{\ell_x}$ of the square is $\color{maroon}{\ell_x}=2\sqrt{1-x}$; and thus, $$A(x)=(\,2\sqrt{1-x}\,)^2.$$ So the volume is $$\int_{0}^1 (\,2\sqrt{1-x}\,)^2 \,dx = \int_{0}^1 4(1-x) \,dx .$$
As can be seen from the plot below, we have a flat triangle with $(0,0),(0,1),(1,0)$ on $xy-$plane. So the required limits here is as $$x|_0^1,y|_0^{1-x},z|_0^{1-x^2}$$
Best Answer
Note that $x^2+y^2-x=(x-\frac12)^2 +y^2-\frac14=0$. Recenter the circle with $x-\frac12=u$ and integrate the volume over the disk $u^2+y^2=\frac14$
$$\int_{u^2+y^2<\frac14} [(1-y^2)-( -2x-3y-10)]dudy \\ =\int_{u^2+y^2<\frac14} (2u+3y-y^2 +12)dudy =\int_{u^2+y^2<\frac14} (12-y^2)dudy \\ =\int_0^{2\pi}\int_0^{1/2} (12-r^2 \sin^2\theta)rdr d\theta =\frac{191\pi}{64}\\ $$