Answer to Question #159414 in Calculus for carrrian

Question

a)    Identify the iterated integrals which are of the following holding Fubini’s Theorem. Give mathematical reason.

i)   ∫_0^ln3▒∫_0^ln2▒〖e^(x+2y) dydx〗                                    iii)  ∫_3^5▒∫_1^2▒〖1/〖(x+y)〗^2 dydx〗

ii)       ∫_(-1)^1▒∫_(〖-x〗^2)^x▒〖(x^2-y)dydx〗                            iv)  ∫_1^(3⁄2)▒∫_y^(3-y)▒ydxdy

a)    The volume in the first octant bounded by the coordinate planes, the plane y = 4, and the

plane  (x/3) + (z/5) = 1.

b)    Sketch the solid in the first octant that is enclosed by the planes x = 0, z = 0, x = 5,

 z − y = 0, and            z = −2y + 6.

Answer 1

Solution

a)

NOTE: Identify the iterated integrals which are of the following holding Fubini’s Theorem. Give mathematical reason.

Fubini's Theorem holds only if a function is continous on a Rectangle $R=[a,b] \times [c,d]$

(i)

$$\intop_0^{ln3} \intop_0^{ln2}e^{x+2y}dydx$$

exponential function will be always continous. Also we have a rectangle

$$[0, ln2]\times[0, ln3]$$

(ii)

$$\intop_{-1}^1\intop_{-x^2}^x (x^2-y)dydx$$

The given case does not hold Fubini's Theorem since limits of y contain variable parts also. It should be continuos on the rectangle area.

(iii)

$$\intop_{3}^5\intop_{1}^2\frac{1}{(x+y)^2}dydx$$

The given function will be discontinous if the denominator equals to zero

$$(x+y^2)=0\\ x+y=0\\ x=-y$$

but for the given limits, x and y will not be negative. Thus x=-y is not possible and so it is continous on Rectangle.

(iv) Similar to the previous case, limits of x contain variable part, thus it does not hold Fubini's Theorem.

Hence

$\implies$ (i), (iii) holds Fubini's Theorem, (ii), (iv) does not hold Fubini's Theorem

a)    The volume in the first octant bounded by the coordinate planes, the plane y = 4, and the

plane  (x/3) + (z/5) = 1.

$\frac z5=(1-\frac x3) \implies z=(\frac{15-5x}{3})\\ \therefore V= \iiint_{E}\delta v = \intop_{y=0}^4 \intop_{x=0}^3 \intop_{0}^{\frac{15-5x}{3}}\delta z \delta x \delta y\\ \therefore V= \intop_{0}^4 \intop_{0}^3 (\frac{15-5x}{3})\ \delta x \delta y = \intop_{0}^4 (\intop_{0}^3 \frac{15-5x}{3}|\ \delta x \delta y\\ = \intop_{0}^4 5x -\frac{5}{3}(\frac{x^2}{2})|_0^3\ \delta y = \intop_{0}^4 (15 -\frac{45}{6})\ \delta y\\ = (15 -\frac{45}{6})(4-0)=\frac{15}{2}(4)= 30\ units^3$

b)    Sketch the solid in the first octant that is enclosed by the planes x = 0, z = 0, x = 5,

 z − y = 0, and           z = −2y + 6.