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We intend to find the volume of a solid described as follows:

The $X$, $Y$ and $Z$ axes are such that the base of the solid is in the $XY$-plane and the vertical direction is parallel to the $Z$-axis. The base of the solid is the region in the $XY$-plane bounded by $y^2$ $=$ $1-x$ and the $Y$-axis. Each cross-section of the solid perpendicular to the $X$-axis is a square.

What would the be the definite integral to calculate the volume of this solid?

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1 Answer 1

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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 $$\int_{0}^1 (\,2\sqrt{1-x}\,)^2 \,dx = \int_{0}^1 4(1-x) \,dx .$$

In the diagram below, we are staring down the $z$-axis. The base of the solid is enclosed by the green curve. A cross section at $\color{maroon} x$ would appear as the line segment $\color{maroon}{\ell_x}$. This cross section is a square with side length $\color{maroon}{\ell_x}=2\color{maroon}{\sqrt{1-x}}$.

enter image description here

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I'm sorry, I made a mistake. I meant $y^2$ $=$ $1-x$ and not $y$ $=$ $1-x^2$. I've edited the question now. –  Sidharth Iyer Feb 25 '12 at 16:36
@SidharthIyer I updated my answer. –  David Mitra Feb 25 '12 at 17:21

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