Can someone help me, thank you

Question

Find the volume of the solid that lies between the surface z=9-x² and xy-plane and bounded on the sides by the planes y=0 and y=4.

Answer 1

To find the volume of the solid that lies between the surface z=9-x^2 and the xy-plane, we can use a triple integral.

Since the solid is bounded on the sides by the planes y=0 and y=4, we can integrate with respect to x, y, and z over the following limits:

0 ≤ y ≤ 4

0 ≤ x ≤ √(9-z)

0 ≤ z ≤ 9

The integral for the volume is:



V = ∫∫∫ dV = ∫0^9 ∫0^√(9-z) ∫0^4 dz dxdy

Integrating with respect to z first, we get:

V = ∫0^4 ∫0^√(9-x^2) 9-x^2 dxdy

Using a trigonometric substitution, let x = 3sinθ, dx = 3cosθdθ, and √(9-x^2) = 3cosθ.

Then the integral becomes:

V = ∫0^π/2 ∫0^3cosθ (9 - 9sin^2θ)3cosθ dxdθ

= 27 ∫0^π/2 ∫0^3cosθ (cos^2θ - sin^2θ) dxdθ

= 27 ∫0^π/2 (sinθ cos^4θ + cos^2θ sin^3θ) dθ

= 27/5