Find The Triple Integral

Find the triple integral of the function f(x,y,z)= x^3cos(y+z) over the cube 2≤x≤5, 0≤y≤π, 0≤z≤π?

I=∫∫∫ f(x,y,z)dxdydz
=∫ x^3dx.∫∫ cos(y+z)dydz

∫ x^3dx=[x^4/4]=1/4(5^4-2^4)=609/4

∫∫ cos(y+z)dydz=∫ [sin(y+z)]dz
=∫ (sin(π+z)-sinz)dz
=∫ (-sinz-sinz)dz
=∫ -2sinzdz
=-2∫ sinzdz
=-2[-cosz]
=2[cosz]
=2(-1-1)
=2(-2)
=-4

I=-4.609/4

I=-609

Find the triple integral of the function f(x,y,z) over the region W, if f(x,y,z)=x2+2y2+z?

∫(x = 0 to 1) ∫(y = -1 to 1) ∫(z = 0 to 2) (x^2 + 2y^2 + z) dz dy dx
= ∫(x = 0 to 1) ∫(y = -1 to 1) ((x^2 + 2y^2)z + z^2/2) {for z = 0 to 2} dy dx
= ∫(x = 0 to 1) ∫(y = -1 to 1) (2x^2 + 4y^2 + 2) dy dx
= 2 ∫(x = 0 to 1) ∫(y = 0 to 1) (2x^2 + 4y^2 + 2) dy dx, since the integrand in y is even
= 2 ∫(x = 0 to 1) (2x^2 y + 4y^3/3 + 2y) {for y = 0 to 1} dx
= 2 ∫(x = 0 to 1) (2x^2 + 10/3) dx
= 2(2x^3/3 + 10x/3) {for x = 0 to 1}
= 8.

I hope this helps!

Triple integral to find the volume of the solid bounded by the parabolic cylinder y=3x^2?

z-bounds: z = 0 to z = 7.

This region projects onto the xy-plane as the region bounded between y = 3x^2 and y = 5
==> y = 3x^2 to y = 5 for x in [-√(5/3), √(5/3)].

So, the volume ∫∫∫ 1 dV equals
∫(x = -√(5/3) to √(5/3)) ∫(y = 3x^2 to 5) ∫(z = 0 to 7) 1 dz dy dx
= ∫(x = -√(5/3) to √(5/3)) ∫(y = 3x^2 to 5) 7 dy dx
= ∫(x = -√(5/3) to √(5/3)) 7(5 - 3x^2) dx
= 2 ∫(x = 0 to √(5/3)) 7(5 - 3x^2) dx, since the integrand is even
= 14(5x - x^3) {for {x = 0 to √(5/3)}
= (140/3) √(5/3).

I hope this helps!

Use the triple integral to find the volume of the given solid. The solid enclosed by the cylinder ....?

You can integrate using variables x, y, and z or using cylindrical coordinates

Method 1: Using variables x, y, z

Cylinder has equation x² + y² = 9
Limits for x are x = -3 to 3
Limits for y are y =-√(9-x²) to √(9-x²)

Cylinder is bounded by the planes y + z = 13 and z = 1
Limits for z are z = 1 to 13 - y

V = ∫{x=-3 to 3} ∫{y = -√(9-x²) to √(9-x²)} ∫{z = 1 to 13-y} dz dy dx
V = ∫{x=-3 to 3} ∫{y = -√(9-x²) to √(9-x²)} z |{z = 1 to 13-y} dy dx
V = ∫{x=-3 to 3} ∫{y = -√(9-x²) to √(9-x²)} ((13-y) - 1) dy dx
V = ∫{x=-3 to 3} ∫{y = -√(9-x²) to √(9-x²)} (12 - y) dy dx
V = ∫{x=-3 to 3} (12y - y²/2) |{y = -√(9-x²) to √(9-x²)} dx
V = ∫{x=-3 to 3} (12(√(9-x²) + √(9-x²)) - 1/2 ((9-x²) - (9-x²))) dx
V = ∫{x=-3 to 3} (24 √(9-x²)) dx

Now we can solve this integral using a trig substitution:
x = 3 sin u .......... dx = 3 cos u du
When x = -3, sin u = -1, u = -π/2
When x = 3, sin u = 1, u = π/2

V = 24 ∫{u=-π/2 to π/2} √(9-9sin²u) * 3 cos u du
V = 24 ∫{u=-π/2 to π/2} 3√(1-sin²u) * 3 cos u du
V = 24 ∫{u=-π/2 to π/2} 3 cos u * 3 cos u du
V = 24 ∫{u=-π/2 to π/2} 9 cos²u du
V = 24 ∫{u=-π/2 to π/2} 9 * 1/2 (1 + cos(2u)) du
V = 24 * 9/2 (u + 1/2 sin(2u)) |{u=-π/2 to π/2}
V = 108 ((π/2 + 1/2 sin(π)) - (-π/2 + 1/2 sin(-π)))
V = 108 (π/2 + π/2)
V = 108 π

--------------------

Method 2: Cylindrical coordinates:

x = r cos θ
y = r sin θ

0 ≤ θ ≤ 2π
0 ≤ r ≤ 3

1 ≤ z ≤ 13 - y
1 ≤ z ≤ 13 - r sin θ

Volume using cylindrical coordinates:

V = ∫{θ=0 to 2π} ∫{r=0 to 3} ∫{z=1 to 13-rsinθ) r dz dr dθ
V = ∫{θ=0 to 2π} ∫{r=0 to 3} rz |{z=1 to 13-rsinθ) dr dθ
V = ∫{θ=0 to 2π} ∫{r=0 to 3} r(13-rsinθ) - r(1) dr dθ
V = ∫{θ=0 to 2π} ∫{r=0 to 3} (12r - r² sinθ) dr dθ
V = ∫{θ=0 to 2π} (6r² - 1/3 r³ sinθ) |{r=0 to 3} dθ
V = ∫{θ=0 to 2π} (54 - 9 sinθ) dθ
V = (54θ + 9 cosθ) |{θ=0 to 2π}
V = (54(2π) + 9(1)) - (0 - 9(1))
V = 108π

Find the triple integral: The solid enclosed by the cylinder y=x^2 and the planes z=o and y+z=1?

From y+ z =1

then y= 1 -z

When z= 0 then y = y1 =1

y will vary from y =x^2(lower) to y= 1(upper)

(Note : Keep track of the limits of integration
otherwise you will have negative answer )

Note that y=x^2 then x can take negative values.

Solve 1= x^2 ==> x = -1(lower ) to 1 (upper)

By the way, answer to this triple integral is 8/15