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Parametric area

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1

For the curve defined by x(t)=2t,y(t)=t2x(t)=2t, y(t)=t^2 from t=0t=0 to 33, what is the area under the curve?

A.18\,unit^2
B.36\,unit^2
C.172\,unit^2
D.15\,unit^2
E.None
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2

Find the area of one arch of the cycloid x=θ−sin⁡θx=\theta-\sin\theta, y=1−cos⁡θy=1-\cos\theta, between θ=0\theta=0 and θ=2π\theta=2\pi.

A.3π3\pi
B.2π2\pi
C.π\pi
D.4π4\pi
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3

For a parametric curve x=x(t)x=x(t), y=y(t)y=y(t), why is the area formula ∫y dx\int y\,dx rewritten as ∫y dxdt dt\int y\,\dfrac{dx}{dt}\,dt rather than integrated directly in xx?

A.Because the curve is described in terms of the parameter tt, not xx directly, so dxdx must be expressed via the chain rule as dxdt dt\frac{dx}{dt}\,dt before integrating over tt
B.Because ∫y dx\int y\,dx is only valid for circles and ellipses
C.Because parametric curves never have an explicit y=f(x)y=f(x) form, so integration in xx is mathematically undefined
D.Because dxdt\frac{dx}{dt} is always equal to 1 for any parametrization, making the substitution trivial
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