Final answer:
The integral ∫∫ᵛ x d A can be expressed as an iterated integral in polar coordinates with x = r cos(θ) and dA = r dr dθ. The region R is the first quadrant bounded by the circle with radius 10 and the line y=x, leading to θ limits of 0 to π/4 and r limits of 0 to 10. Therefore, the integral in polar coordinates becomes ∫θ(0 to π/4) ∫r(0 to 10) r² cos(θ) dr dθ.
Step-by-step explanation:
To express the integral ∫∫ᵛ x d A as an iterated integral using polar coordinates over the region R bounded by y=√(100-x²), the positive x-axis, and the line y=x, we need to represent x in terms of polar coordinates, which is x = r · cos(θ). Next, we convert the area differential dA in polar coordinates to r dr dθ. The integration limits for r and θ must correspond to the region R. The region lies in the first quadrant bounded by the circle radius 10, since y=√(100-x²) is the upper half of a circle centered at the origin with radius 10.
The boundary y=x represents a line with a 45-degree angle, which corresponds to θ = π/4 in polar coordinates. As the integral over the region will be symmetric along this line, the bounds for θ will be from 0 to π/4. For r, since the circle's equation is x² + y² = 10², in polar coordinates this becomes r² = 100, which simplifies to r = 10. Therefore, the limits for r are from 0 to 10.
Combining this together, the iterated integral becomes:
θ₁∫^{θ₂} ᵛ₁∫ʹ ᵛ r · cos(θ) r dr dθ,
with f(r,θ)= r · cos(θ) · r = r^2 · cos(θ).
The limits for θ are 0 to π/4 and for r are 0 to 10. The iterated integral is:
0∫^{π/4} 0∫^{10} r^2 · cos(θ) dr dθ.