Math  /  Calculus

QuestionEvaluate the line integral, where CC is the given curve. Cxyz2ds,C\int_{C} x y z^{2} d s, C is the line segment from (2,3,0)(-2,3,0) to (0,4,4)(0,4,4)

Studdy Solution
Evaluate the integral:
First, simplify the expression inside the integral: (2+2t)(3+t)(16t2)=(2+2t)(48t2+16t3) (-2 + 2t)(3 + t)(16t^2) = (-2 + 2t)(48t^2 + 16t^3)
Expand and simplify: =(96t232t3+96t3+32t4) = (-96t^2 - 32t^3 + 96t^3 + 32t^4) =(96t2+64t3+32t4) = (-96t^2 + 64t^3 + 32t^4)
Now, integrate: 01(96t2+64t3+32t4)21dt \int_{0}^{1} (-96t^2 + 64t^3 + 32t^4) \sqrt{21} \, dt
=21[01(96t2)dt+01(64t3)dt+01(32t4)dt] = \sqrt{21} \left[ \int_{0}^{1} (-96t^2) \, dt + \int_{0}^{1} (64t^3) \, dt + \int_{0}^{1} (32t^4) \, dt \right]
Calculate each integral: 01(96t2)dt=[32t3]01=32 \int_{0}^{1} (-96t^2) \, dt = \left[ -32t^3 \right]_{0}^{1} = -32 01(64t3)dt=[16t4]01=16 \int_{0}^{1} (64t^3) \, dt = \left[ 16t^4 \right]_{0}^{1} = 16 01(32t4)dt=[325t5]01=325 \int_{0}^{1} (32t^4) \, dt = \left[ \frac{32}{5}t^5 \right]_{0}^{1} = \frac{32}{5}
Combine the results: 21(32+16+325)=21(16+325) \sqrt{21} \left( -32 + 16 + \frac{32}{5} \right) = \sqrt{21} \left( -16 + \frac{32}{5} \right)
=21(805+325) = \sqrt{21} \left( -\frac{80}{5} + \frac{32}{5} \right) =21(485) = \sqrt{21} \left( -\frac{48}{5} \right) =48215 = -\frac{48\sqrt{21}}{5}
The value of the line integral is:
48215 \boxed{-\frac{48\sqrt{21}}{5}}

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