# Find the volume of the solid bounded by the paraboloids z=5(x^2)+5(y^2) and z=6-7(x^2)-(y^2).

### 1 Answer | Add Yours

The given two surfaces interact on the elliptec cylinder 5x^2+5y^2 = z = 6-7x^2-y^2 that is,

(5x^2 +5y^2 +7x^2+y^2 = 6 , => 12x^2+6y^2 = 6, dividing through out by 6 , we get 2x^2+y^2 = 1)

The orthogonal projection of the volume on the xy-plane is the plane region R enclosed by the ellipse 2x^2+y^2 = 1............(1),

which determines the limits for x and y

z: from 5x^2+5y^2 to 6-7x^2-y^2

y: from -`sqrt(1-2x^2)` to `sqrt(1-2x^2)` (solving for y in (1))

x: from -1/`sqrt(2)` to +1/`sqrt(2)` (taking y = 0 in (1))

therefore required volume = ```int_(-1/sqrt(2))^(+1/sqrt(2))` `int_(-sqrt(1-2x^2))^(sqrt(1-2x^2))`

` ` `int_(5x^2+5y^2)^(6-7x^2-y^2)` dzdydx

= `int_(-1/sqrt(2))^(1/sqrt(2))` `int_(-sqrt(1-2x^2))^(sqrt(1-2x^2))` ((6-7x^2-y^2)-(5x^2+5y^2))dy.dx

= `int_(-1/sqrt(2))^(1/sqrt(2))` `int_(-sqrt(1-2x^2))^(sqrt(1-2x^2))` (6-12x^2-6y^2)dy.dx (taking 6 common)

= 6 `int_(-1/sqrt(2))^(1/sqrt(2))` `int_(-sqrt(1-2x^2))^(sqrt(1-2x^2))` (1-2x^2-y^2) dydx

= 12`int_(-1/sqrt(2))^(1/sqrt(2))` `int_(0)^(sqrt(1-2x^2))` ((1-2x^2) - y^2) dydx , (changing the lower limit to 0)

= 12 `int_(-1/sqrt(2))^(1/sqrt(2))` [ (1-2x^2)y - y^3/3 ]dx , with the limits 0 to`sqrt(1-2x^2)` ,(apply the limits)

= 12 `int_(-1/sqrt(2))^(1/sqrt(2))`( (1-2x^2)(`sqrt(1-2x^2)` - (`sqrt(1-2x^2)` )^3 `/` 3 - 0) dx

= 12`int_(-1/sqrt(2))^(1/sqrt(2))`( (1-2x^2)^3/2 - (1-2x^2)^3 `/` 3 )dx

= 12`int_(-1/sqrt(2))^(1/sqrt(2))` (1-2x^2)^3/2 ( 1-1/3) dx

= 12`int_(-1/sqrt(2))^(1/sqrt(2))` 2(1-2x^2)^3/2 `/3` dx

= 8`int_(-1/sqrt(2))^(1/sqrt(2))` (1-2x^2)^3/2 dx (making the lower limit 0)

= 16`int_0^(1/sqrt(2))` (1-2x^2)^3/2 dx,

to evaluate this integral let us suppose that x = 1/`sqrt(2)` sin`theta` , such that dx = 1/`sqrt(2)` cos`theta` d`theta`

and for x=1/`sqrt(2)` , sin`theta` = 1 , such that `theta` = pi/2,

so the integral becomes

= 16`int_0^(pi/2)'` (1-2(1/`sqrt(2)sintheta` )^2)^3/2 * 1/`sqrt(2)` cos`theta` d`theta`

= 16`int_0^(pi/2)` (1-sin^2`theta` )^3/2 *1/`sqrt(2)` *cos`theta` d`theta`

= 16/`sqrt(2)` `int_0^(pi/2)` (cos^2`theta` )^3/2*cos`theta` d`theta`

=16/`sqrt(2)` `int_0^(pi/2)` cos^4`theta` d`theta` = 16/`sqrt(2)` (3/4*1/2*pi/2) = 3pi/`sqrt(2)`

thus the volume is 3pi/`sqrt(2)`