Finding the volume of the solid

Question

1. I have a solid $E$ below that I would like to compute the volume. I can slice $E$ by cross-sections $E_x$ that are perpendicular to Ox. Then $$ |E|=\int_0^1 E_x \,dx = \int_0^1 \frac{1}{2}z[e^{2x}-(-e^x)] \,dx = \frac{1}{2} \int_0^1 e^{3x} (e^{2x} + e^x) \,dx = \frac{4e^5+5e^4-9}{40}. $$ Am I right?

2. I want to do a different way. I divide the solid into two parts: $y<0$ (left solid) and $y>0$ (right solid). The volume for left and right solids is given by the integral of the equation of the left and right side surfaces. However, I don't know how to find the equations of these surfaces. Is there a way to do it?

Thank you.

Answer 1

For the part of the solid where $y>0$ (the blue "half" in the image below), one can parameterize the surface by

$$\vec r(s,t) = (1-t) \left\langle s,0,e^{3s}\right\rangle + t \left\langle s,e^{2s},0\right\rangle$$

with $(s,t)\in[0,1]^2$. In other words,

$$\begin{cases}x = s \\ y = te^{2s} \\ z=(1-t)e^{3s}\end{cases}$$

and eliminating the parameters yields

$$x=s \implies y=te^{2x} \implies z=\left(1-ye^{-2x}\right) e^{3x}$$

For $y<0$ (red), the only difference is in the $y$-coordinate of the parameterization,

$$\vec r(s,t) = (1-t) \left\langle s,0,e^{3s}\right\rangle + t \left\langle s,-e^{s},0\right\rangle$$

so that

$$x=s \implies y=-te^x \implies z=\left(1+ye^{-x}\right) e^{3x}$$

Answer 2

The first part of your question is correct. The differential volume of a representative slice of the solid at some point $x \in [0,1]$ is $$dV = \frac{1}{2} b(x) h(x) \, dx = \frac{1}{2} (e^2x - (-e^x)) e^{3x} \ , dx,$$ where $b(x)$ is the width of the triangular base as a function of $x$, and $h(x)$ is the height of the triangle.

For the second part of your question, for some value $x$, the equation of the line joining the points $(x, e^{2x}, 0)$ and $(x, 0, e^{3x})$, is given by the point-slope formula for a line: $$z - 0 = \frac{e^{3x} - 0}{0 - e^{2x}} (y - e^{2x})$$ or equivalently $$z = e^x (e^{2x} - y).$$ So for a point $(x,y)$ such that $x \in [0,1]$ and $y \in [0, e^{2x}]$, the above equation gives the height. Similarly, the line joining the points $(x, -e^x, 0)$ and $(x, 0, e^{3x})$ is $$z - 0 = \frac{e^{3x} - 0}{0 - (-e^x)}(y - (-e^x))$$ or $$z = e^{2x}(y + e^x).$$ So the full function for the height of the surface is $$z(x,y) = \begin{cases} e^x (e^{2x} - y), & x \in [0,1], y \in [0, e^{2x}] \\ e^{2x}(y + e^x), & x \in [0,1], y \in [-e^x, 0). \end{cases} $$ Then the volume is expressed in terms of the above by $$V = \int_{x=0}^1 \int_{y=-e^x}^{e^{2x}} z(x,y) \, dy \, dx.$$