Evaluate C∫ (2x2 + 3y2)dx, where C is the curve given by x(t)=at2 , y(t)=2at, 0≤t≤1.
Let us evaluate ∫C(2x2+3y2)dx\int_C (2x^2 + 3y^2)dx∫C(2x2+3y2)dx, where CCC is the curve given by x(t)=at2,y(t)=2at,0≤t≤1.x(t)=at^2 , y(t)=2at, 0≤t≤1.x(t)=at2,y(t)=2at,0≤t≤1.
∫C(2x2+3y2)dx=∫01(2(at2)2+3(2at)2)d(at2)=∫01(2a2t4+12a2t2)2atdt=∫01(4a3t5+24a3t3)dt=4a3∫01(t5+6t3)dt=4a3(t66+6t44)∣01=4a3(16+32)=4a3106=20a33.\int_C (2x^2 + 3y^2)dx=\int_0^1 (2(at^2)^2 + 3(2at)^2)d(at^2)=\int_0^1 (2a^2t^4 + 12a^2t^2)2atdt= \int_0^1 (4a^3t^5 + 24a^3t^3)dt=4a^3\int_0^1 (t^5 + 6t^3)dt=4a^3(\frac{t^6}{6}+6\frac{t^4}{4})|_0^1= 4a^3(\frac{1}{6}+\frac{3}{2})=4a^3\frac{10}{6}=\frac{20a^3}{3}.∫C(2x2+3y2)dx=∫01(2(at2)2+3(2at)2)d(at2)=∫01(2a2t4+12a2t2)2atdt=∫01(4a3t5+24a3t3)dt=4a3∫01(t5+6t3)dt=4a3(6t6+64t4)∣01=4a3(61+23)=4a3610=320a3.
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