Respuesta :
Answer:
Required value is [tex]\frac{1250}{3}[/tex].
Step-by-step explanation:
We have to find the value of,
[tex]\int\int_{D} x^2y dA\hfill (1)[/tex]
where, [tex]D=\{(5,\theta) : 0\leq r\leq 5, 0\leq \theta \leq \pi \}[/tex]
To convert it in polar form let [tex]x=r\cos\theta, y=r\sin\theta[/tex] and substitute in (1) we get,
[tex]\int\int_{D} x^2y dA[/tex]
[tex]=\int_{0}^{\pi}\int_{0}^{5}(r\cos\theta)^2(r\sin\theta)rdrd\theta[/tex]
Assuming,
[tex]u=\cos \theta[/tex] then,
[tex]du=-\sin\theta d\theta\implies-du=\sin\thetad\theta[tex]
For [tex]\theta=0, u=1[/tex] and [tex]\theta=\pi[/tex] gives u=-1. Hence,
[tex]\int\int_{D} x^2y dA[/tex]
[tex]=\int_{0}^{\pi}\int_{0}^{5}(r\cos\theta)^2(r\sin\theta)rdrd\theta[/tex]
[tex]=\int_{1}^{-1}u^2(-du)\int_{0}^{5}r^4dr[/tex]
[tex]=\int_{-1}^{1}u^2du\int_{0}^{5}r^4dr[/tex]
[tex]=\Big[\frac{u^3}{3}\Big]_{-1}^{1}\times\Big[\frac{r^5}{5}\Big]_{0}^{5}[/tex]
[tex]=\frac{2\times 5^4}{3}=\frac{1250}{3}[/tex]
The change to polar coordinates should be done according to:
x = r×cosθ y = r× sinθ dA = r×dr×dθ
The solution is:
∫∫ x²×y×dA = 5000×D⁵/ 3
∫∫ x²×y×dA ⇒ ∫∫ [(r×cosθ)²× r×sinθ]× r×dr×dθ
∫∫ r²×cos²θ×r×sinθ× r×dr×dθ
Integration limits:
x² + y² = r² ⇒ r²×cos²θ + r²× sin²θ = (5D)²
r² × ( cos²θ + sin²θ ) = (5D)²
r² = (5D)² ⇒ r = 5×D
Then integration limits are:
0 ≤ r ≤ 5×D
∫∫ r²×cos²θ×r×sinθ× r×dr×dθ = ∫ r³ dr ×∫cos²θ×r×sinθ×dθ ( 0 ≤ θ ≤ π )
∫ r³ dr = r⁴/4|₀⁵D ⇒ (625×D⁵) / 4
∫cos²θ×r×sinθ×dθ we change variables u = cosθ du = - sinθ×dθ
- ∫ u²× du = - u³/3 |₁⁻¹ when θ = 0 cos0 = 1
when θ = π cos π = -1
- [ - 1/3 - ( 1/3)] = - ( -2/3 ) = (2/3)
- ∫ u²× du = 2/3
∫∫ x²×y×dA = (625×D⁵) / 4×(2/3)
∫∫ x²×y×dA = 5000×D⁵/ 3
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