Answer:
The volume is [tex]W = 668[/tex]
Step-by-step explanation:
From the question we are told that
The region bounded by [tex]z = 100- y^2 , y = 10x^2[/tex]
The z plane is = 0
The volume of this region is mathematically evaluated as
[tex]V = \int\limits^b_a \int\limits^d_c \int\limits^f_e {dx dy dz} \,[/tex]
Now we need to obtain a,b,c,d,e,f
So from the question we are told that [tex]z = 1 00 - y^2[/tex]
So
[tex]0 \le z \le 100-y^2[/tex]
We also told on plane where the region is that z = 0 this implies that it an x-y plane so
[tex]0 = 100 - y^2[/tex]
=> [tex]y^2 = 100[/tex]
[tex]y = 10[/tex]
On the y- axis the second equation bounding the region is [tex]y = 10x^2[/tex]
So
[tex]10x^2 \le y \le 10[/tex]
Now the point where this two equation ([tex]y =10x^2 \ and \ y = 10[/tex]) intersect is
mathematically represented as
[tex]10x^2 = 10[/tex]
[tex]x = \pm 1[/tex]
So
[tex]-1 \le x \le 1[/tex]
Hence a = - 1
b = 1
c = [tex]10x^2[/tex]
d = 10
e= 0
f = [tex]100 -y^2[/tex]
So
[tex]W = \int\limits^{1}_{-1} \int\limits^{10}_{10x^2} \int\limits^{100 -y^2}_0 {dx dy dz} \,[/tex]
[tex]W= \int\limits^{1}_{-1} \int\limits^{10}_{10x^2} [z] \left | {100-y^2} \atop {0}} \right. {dx dy } \,[/tex]
[tex]W = \int\limits^{1}_{-1} \int\limits^{10}_{10x^2} [100 -y^2] {dx dy } \,[/tex]
[tex]W = \int\limits^{1}_{-1} [100y - \frac{y^3}{3} ] \left | 10} \atop {10x^2}} \right. {dx } \,[/tex]
[tex]W = \int\limits^{1}_{-1} [([100(10) ]- 100(10x^2) ) - [ (\frac{10^3}{3}) - (\frac{(10x^2)^3}{3} ) ]] {dx } \,[/tex]
[tex]W = \int\limits^{1}_{-1} [([1000 - 1000x^2 ) - [ (\frac{10^3}{3}) - (\frac{(10x^2)^3}{3} ) ]] {dx } \,[/tex]
[tex]W = [([1000x - \frac{1000x^3}{3} ) - [ (\frac{10^3}{3}) x- (\frac{(10x^7)}{21} ) ]] \left | 1} \atop {-1}} \right.[/tex]
[tex]W = [([1000(1) - \frac{1000(1)^3}{3} ) - [ (\frac{10^3}{3}) (1)- (\frac{(10(1)^7)}{21} ) ]]-[/tex]
[tex][([1000(-1) - \frac{1000(-1)^3}{3} ) - [ (\frac{10^3}{3}) (-1)- (\frac{(10(-1)^7)}{21} ) ]][/tex]
[tex]W = 667 - 333 - (- 667 + 333 )[/tex]
[tex]W = 668[/tex]
