Double Integral, Exercise 3

Question

Use a double integral to calculate the volume enclosed between the partial surface of the elliptic paraboloid $z = h (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) \geq 0$ and the (x,y) plane, where h is a positive constant!

Express h in terms of c, if this volume equals the volume enclosed between the partial surface of the ellipsoid $z = c \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \geq 0$ 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. ...

$\int \int f (x, y) d x d y = \int_{y_{\min}}^{y_{\max}} d y \int_{x_{\min} (y)}^{x_{\max} (y)} f (x, y) d x$	(3.2.1.7)

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

$\int \int f (x, y) d x d y = \int_{x_{\min}}^{x_{\max}} d x \int_{y_{\min} (x)}^{y_{\max} (x)} f (x, y) d y$	(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),

${\begin{cases} \frac{V}{2} = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} z (x, y) d y = c \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \\ (substitution: \sin u = \frac{\frac{y}{b}}{\sqrt{1 - \frac{x^{2}}{a^{2}}}}, d y = b \sqrt{1 - \frac{x^{2}}{a^{2}}} \cos u d u) \\ \frac{V}{2} = c b \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u = \frac{c b π}{2} \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) = \\ = \frac{π b c}{2} (x - \frac{x^{3}}{3 a^{2}}) \|_{- a}^{a} = \frac{π b c}{2} \frac{4 a}{3} \Rightarrow V = \frac{4 π}{3} a b c \end{cases}}$	(3.2.1.26)

{\begin{cases} \frac{V}{2} = \int_{x_{1}}^{x_{2}} d x \int_{y_{1} (x)}^{y_{2} (x)} z (x, y) d y = c \int_{- a}^{a} d x \int_{- \frac{b}{a} \sqrt{a^{2} - x^{2}}}^{\frac{b}{a} \sqrt{a^{2} - x^{2}}} d y \sqrt{1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}} \\ (substitution: \sin u = \frac{\frac{y}{b}}{\sqrt{1 - \frac{x^{2}}{a^{2}}}}, d y = b \sqrt{1 - \frac{x^{2}}{a^{2}}} \cos u d u) \\ \frac{V}{2} = c b \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) \int_{- \frac{π}{2}}^{\frac{π}{2}} \cos^{2} u d u = \frac{c b π}{2} \int_{- a}^{a} d x (1 - \frac{x^{2}}{a^{2}}) = \\ = \frac{π b c}{2} (x - \frac{x^{3}}{3 a^{2}}) |_{- a}^{a} = \frac{π b c}{2} \frac{4 a}{3} \Rightarrow V = \frac{4 π}{3} a b c \end{cases}}

(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:

$x \Leftrightarrow y and a \Leftrightarrow b$

Parts 1-4

Solution of question 1.

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

${\begin{cases} V = h \int_{- a}^{a} d x \int_{y {(x)}_{\min}}^{y {(x)}_{\max}} (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}) d y \\ where {\begin{cases} y {(x)}_{\min} = - \frac{b}{a} \sqrt{a^{2} - x^{2}} \\ y {(x)}_{\max} = \frac{b}{a} \sqrt{a^{2} - x^{2}} \end{cases}} \end{cases}$

As a consequence, the first integration gives

$\begin{array}{l} V = h \int_{- a}^{a} d x [y (1 - \frac{x^{2}}{a^{2}} - \frac{y^{2}}{3 b^{2}})] |_{y {(x)}_{\min}}^{y {(x)}_{\max}} = \\ = h \int_{- a}^{a} d x {[(y {(x)}_{\max} - y {(x)}_{\min}) (1 - \frac{x^{2}}{a^{2}})] - \frac{y {(x)}_{\max}^{3} - y {(x)}_{\min}^{3}}{3 b^{2}}} = \\ = h \int_{- a}^{a} d x {\frac{2 b}{a^{3}} {\sqrt{a^{2} - x^{2}}}^{3} - \frac{2 b^{3}}{a^{3} 3 b^{2}} {\sqrt{a^{2} - x^{2}}}^{3}} = \\ = \frac{2 b h}{a^{3}} \int_{- a}^{a} d x (1 - \frac{1}{3}) {\sqrt{a^{2} - x^{2}}}^{3} = \frac{4 b h}{3 a^{3}} \int_{- a}^{a} d x {\sqrt{a^{2} - x^{2}}}^{3} \end{array}$

From the table of indefinite integrals, we have:

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

which after substitution in the result of part 2, give

$V = \frac{4 b h}{3 a^{3}} \frac{3 a^{4}}{8} [asin (1) - asin (- 1)] = \frac{π a b h}{2}$

Part 5

Solution of question 2.

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

$\frac{1}{2} π a b h = \frac{2}{3} π a b c$ 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.

Math Animated™ Mathematical Introduction for Physics and Engineering by Samuel Dagan (Copyright © 2007-2020)

Chapter 3: Many Variables; Section 2: Integration; page 1