Question #204868

Evaluate C∫ (2x2 + 3y2)dx, where C is the curve given by x(t)=at2 , y(t)=2at, 0≤t≤1.


1
Expert's answer
2021-06-09T15:10:30-0400

Let us evaluate C(2x2+3y2)dx\int_C (2x^2 + 3y^2)dx, where CC is the curve given by x(t)=at2,y(t)=2at,0t1.x(t)=at^2 , 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=4a301(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}.



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