]> Exercise 3

# Double Integral, Exercise 3

## Question

1. Use a double integral to calculate the volume enclosed between the partial surface of the elliptic paraboloid $z=h\left(1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}\right)\ge 0$ and the (x,y) plane, where h is a positive constant!
2. Express h in terms of c, if this volume equals the volume enclosed between the partial surface of the ellipsoid $z=c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}\ge 0$ and the (x,y) plane, while a and b keep the same values!

## Reminder

We evaluate a double integral, by iteration of two single integrals, first on x and then on y, or in reverse order. Since the evaluation of a single integral is a known procedure, we have only to learn how to define the limits of the iterated integrals. If the first iteration is along x, then the limits of the integral are y dependent, according to the boundary of the region of integration, and therefore the value of the first iteration is a function of y. The second iteration is a regular single integral, with limits of constant values.

The integration over a convex region will be used, in order to clarify this. ...

 $\int \int f\left(x,y\right)\text{d}x\text{d}y=\underset{{y}_{\mathrm{min}}}{\overset{{y}_{\mathrm{max}}}{\int }}\text{d}y\underset{{x}_{\mathrm{min}}\left(y\right)}{\overset{{x}_{\mathrm{max}}\left(y\right)}{\int }}f\left(x,y\right)\text{d}x$ (3.2.1.7)

If the first integration would be done with respect to y, ....

 $\int \int f\left(x,y\right)\text{d}x\text{d}y=\underset{{x}_{\mathrm{min}}}{\overset{{x}_{\mathrm{max}}}{\int }}\text{d}x\underset{{y}_{\mathrm{min}}\left(x\right)}{\overset{{y}_{\mathrm{max}}\left(x\right)}{\int }}f\left(x,y\right)\text{d}y$ (3.2.1.8)

In both cases, (3.2.1.7) and (3.2.1.8), first one has to evaluate the inner-most integral. .....

..... a double integral of a surface z(x,y), over a (x,y) region of the variables, represents the "volume" under this surface.

..... we'll calculate the volume of the ellipsoid

 $\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}+\frac{{z}^{2}}{{c}^{2}}=1$ (3.2.1.24)

Half of its volume is enclosed between the surface corresponding to positive z

 $z=c\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}$ (3.2.1.25)

and the (x,y) plane (corresponding to z=0), therefore the volume V can be calculated over the same region of coordinates as in (3.2.1.21),

 $\left\{\begin{array}{l}\frac{V}{2}=\underset{{x}_{1}}{\overset{{x}_{2}}{\int }}\text{d}x\underset{{y}_{1}\left(x\right)}{\overset{{y}_{2}\left(x\right)}{\int }}z\left(x,y\right)\text{d}y=c\underset{-a}{\overset{a}{\int }}\text{d}x\underset{-\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}}{\overset{\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}}{\int }}\text{d}y\sqrt{1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}}\\ \left(\text{substitution:}\text{\hspace{0.28em}}\mathrm{sin}u=\frac{\frac{y}{b}}{\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}}\text{\hspace{0.17em}},\text{\hspace{0.28em}}\text{d}y=b\sqrt{1-\frac{{x}^{2}}{{a}^{2}}}\mathrm{cos}u\text{d}u\right)\\ \frac{V}{2}=cb\underset{-a}{\overset{a}{\int }}\text{d}x\left(1-\frac{{x}^{2}}{{a}^{2}}\right)\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}{\mathrm{cos}}^{2}u\text{d}u=\frac{cb\pi }{2}\underset{-a}{\overset{a}{\int }}\text{d}x\left(1-\frac{{x}^{2}}{{a}^{2}}\right)=\\ =\frac{\pi bc}{2}\left(x-\frac{{x}^{3}}{3{a}^{2}}\right)\underset{-a}{\overset{a}{|}}=\frac{\pi bc}{2}\frac{4a}{3}\text{ }⇒\text{ }V=\frac{4\pi }{3}abc\end{array}\right\}$ (3.2.1.26)

## Notice for the solution of question 1

In the solution, the first integration is done over y. It could be done over x. In such a case one just has to invert:

$x⇔y\text{ }\text{and}\text{ }a⇔b$

## Parts 1-4

Solution of question 1.

1. The required volume V is expressed by the following double integral:

$\left\{\begin{array}{l}V=h\underset{-a}{\overset{a}{\int }}\text{d}x\underset{y{\left(x\right)}_{\mathrm{min}}}{\overset{y{\left(x\right)}_{\mathrm{max}}}{\int }}\left(1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}\right)\text{d}y\\ \text{where}\text{\hspace{0.28em}}\left\{\begin{array}{l}y{\left(x\right)}_{\mathrm{min}}=-\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\\ y{\left(x\right)}_{\mathrm{max}}=\frac{b}{a}\sqrt{{a}^{2}-{x}^{2}}\end{array}\right\}\end{array}$

2. As a consequence, the first integration gives

$\begin{array}{l}V=h\underset{-a}{\overset{a}{\int }}\text{d}x\left[y\left(1-\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{3{b}^{2}}\right)\right]\underset{y{\left(x\right)}_{\mathrm{min}}}{\overset{y{\left(x\right)}_{\mathrm{max}}}{|}}=\\ =h\underset{-a}{\overset{a}{\int }}\text{d}x\left\{\left[\left(y{\left(x\right)}_{\mathrm{max}}-y{\left(x\right)}_{\mathrm{min}}\right)\left(1-\frac{{x}^{2}}{{a}^{2}}\right)\right]-\frac{y{\left(x\right)}_{\mathrm{max}}^{3}-y{\left(x\right)}_{\mathrm{min}}^{3}}{3{b}^{2}}\right\}=\\ =h\underset{-a}{\overset{a}{\int }}\text{d}x\left\{\frac{2b}{{a}^{3}}{\sqrt{{a}^{2}-{x}^{2}}}^{3}-\frac{2{b}^{3}}{{a}^{3}3{b}^{2}}{\sqrt{{a}^{2}-{x}^{2}}}^{3}\right\}=\\ =\frac{2bh}{{a}^{3}}\underset{-a}{\overset{a}{\int }}\text{d}x\left(1-\frac{1}{3}\right){\sqrt{{a}^{2}-{x}^{2}}}^{3}=\frac{4bh}{3{a}^{3}}\underset{-a}{\overset{a}{\int }}\text{d}x{\sqrt{{a}^{2}-{x}^{2}}}^{3}\end{array}$

3. From the table of indefinite integrals, we have:

$\int \text{d}x{\sqrt{{a}^{2}-{x}^{2}}}^{3}=\frac{x{\sqrt{{a}^{2}-{x}^{2}}}^{3}}{4}+\frac{3{a}^{2}x\sqrt{{a}^{2}-{x}^{2}}}{8}+\frac{3{a}^{4}}{8}\text{asin}\left(\frac{x}{a}\right)$

4. which after substitution in the result of part 2, give

$V=\frac{4bh}{3{a}^{3}}\frac{3{a}^{4}}{8}\left[\text{asin}\left(1\right)-\text{asin}\left(-1\right)\right]=\frac{\pi abh}{2}$

## Part 5

Solution of question 2.

1. The result of part 4 should be compared to half a volume of the ellipsoid (3.2.1.26)

$\frac{1}{2}\pi abh=\frac{2}{3}\pi abc$ yielding finally $h=\frac{4}{3}c$

## Score

By parts.
Any one of the 5 parts is worth 2 points.

By questions.
Question 1 is worth 8 points.
Question 2 is worth 2 points.