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Question Number 122807 by TITA last updated on 19/Nov/20

Commented by TITA last updated on 19/Nov/20

$p l e a s e h e l p$
	Answered by ebi last updated on 20/Nov/20
	$V=∫∫_E ∫ f(x,y,z) dV volume enclosed by z=x+y, z=0 y=x^2 , x=y^2 the limit of z is 0≤z≤x+y. on xy−plane, the bounded region is between y=x^2 and x=y^2 , thus, the limit of y is x^2 ≤y≤(√x). now, finding the domain of x, x=x^4 ⇒x^4 −x=0⇒x(x^3 −1)=0 x=0 or x=1, thus, limit of x is 0≤x≤1. V=∫∫_D ∫_(u_1 (x,y)) ^(u_2 (x,y)) f(x,y,z) dz dA E={(x,y,z)∣(x,y)∈D, 0≤z≤x+y} D={(x,y)∣0≤x≤1, x^2 ≤y≤(√x)} V=∫_0 ^1 ∫_x^2 ^(√x) ∫_0 ^(x+y) xy dz dy dx V=∫_0 ^1 ∫_x^2 ^(√x) xyz∣_0 ^(x+y) dy dx V=∫_0 ^1 ∫_x^2 ^(√x) x^2 y+xy^2 dy dx V=∫_0 ^1 ((x^2 y^2 )/2)+((xy^3 )/3)∣_x^2 ^(√x) dx V=∫_0 ^1 (x^3 /2)+(x^(5/2) /3)−(x^6 /2)−(x^7 /3) dx V= (x^4 /8)+((2x^(7/2) )/(21))−(x^7 /(14))−(x^8 /(24))∣_0 ^1 V=(1/8)+((2(1))/(21))−(1/(14))−(1/(24))=(3/(28))$
	$V = \int \int_{E} \int f (x, y, z) d V$ $v o l u m e e n c l o s e d b y$ $z = x + y, z = 0$ $y = x^{2}, x = y^{2}$ $t h e l i m i t o f z i s 0 ⩽ z ⩽ x + y .$ $o n x y - p l a n e, t h e b o u n d e d r e g i o n i s$ $b e t w e e n y = x^{2} a n d x = y^{2},$ $t h u s, t h e l i m i t o f y i s x^{2} ⩽ y ⩽ \sqrt{x} .$ $n o w, f i n d i n g t h e d o m a i n o f x,$ $x = x^{4} \Rightarrow x^{4} - x = 0 \Rightarrow x (x^{3} - 1) = 0$ $x = 0 o r x = 1,$ $t h u s, l i m i t o f x i s 0 ⩽ x ⩽ 1 .$ $V = \int \int_{D} \int_{u_{1} (x, y)}^{u_{2} (x, y)} f (x, y, z) d z d A$ $E = {(x, y, z) ∣ (x, y) \in D, 0 ⩽ z ⩽ x + y}$ $D = {(x, y) ∣ 0 ⩽ x ⩽ 1, x^{2} ⩽ y ⩽ \sqrt{x}}$ $V = \underset{0}{\int^{1}} \int_{x^{2}}^{\sqrt{x}} \int_{0}^{x + y} x y d z d y d x$ $V = \underset{0}{\int^{1}} \int_{x^{2}}^{\sqrt{x}} x y z ∣_{0}^{x + y} d y d x$ $V = \underset{0}{\int^{1}} \int_{x^{2}}^{\sqrt{x}} x^{2} y + {x y}^{2} d y d x$ $V = \underset{0}{\int^{1}} \frac{x^{2} y^{2}}{2} + \frac{{x y}^{3}}{3} ∣_{x^{2}}^{\sqrt{x}} d x$ $V = \underset{0}{\int^{1}} \frac{x^{3}}{2} + \frac{x^{\frac{5}{2}}}{3} - \frac{x^{6}}{2} - \frac{x^{7}}{3} d x$ $V = \frac{x^{4}}{8} + \frac{2 x^{\frac{7}{2}}}{21} - \frac{x^{7}}{14} - \frac{x^{8}}{24} ∣_{0}^{1}$ $V = \frac{1}{8} + \frac{2 (1)}{21} - \frac{1}{14} - \frac{1}{24} = \frac{3}{28}$
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