Question

Find the area of the lateral surface (see figure) over the curve \(C\) in the \(x y\) -plane and under the surface \(z=f(x, y),\) where Lateral surface area \(=\int_{C} f(x, y) d s\) \(f(x, y)=y+1, \quad C: y=1-x^{2}\) from (1,0) to (0,1)

Short answer

The area of the lateral surface over the curve \(C\) in the xy-plane and under the surface \(z=f(x, y)\), computed as \(\int_{C} f(x, y) ds\) where \(f(x, y)=y+1\) and \(C: y=1-x^{2}\) from (1,0) to (0,1) is given by the integral \(\int_{0}^{1} 2t^2 \sqrt{1+4(1-t)^2} dt\). The integral can be evaluated numerically for an approximation or it can be left in this form if an exact answer is not required.

Step-by-step solution

Parameterize the curve C

To find the parameterization of \(C: y=1-x^{2}\) from (1,0) to (0,1), one can use the parameter \(t\) and note that \(x\) travels from 1 to 0 as \(t\) ranges over the interval [0,1]. A proper parameterization can be chosen as \(x=1-t\) and \(y=1-(1-t)^2\).

Compute dx/dt and dy/dt

Differentiating \(x\) with respect to \(t\) gives \(\frac{dx}{dt}=-1\), differentiating \(y\) gives \(\frac{dy}{dt}=2(1-t)\), this will later be needed to compute \(ds\).

Calculate the differential of arc length ds

The arc length differential \(ds\) is given by the formula \(\sqrt{dx^2+dy^2}\), but since our parameterization is a function of \(t\), it can be re-expressed as \(\sqrt{(dx/dt)^2 + (dy/dt)^2} dt\). Substituting the derivatives calculated on step 2 we have \(ds=\sqrt{(-1)^2 + [2(1-t)]^2} dt\). Simplifying gives \(ds=\sqrt{1+4(1-t)^2} dt\)

Substitute into the integral

Now substitute \(f(x,y), ds\) and the limits of \(t\) into the integral. The integral becomes \(\int_{0}^{1} (1-(1-t)^2 + 1) \sqrt{1+4(1-t)^2} dt\). Simplifying the integrand transforms the integral to \(\int_{0}^{1} 2t^2 \sqrt{1+4(1-t)^2} dt\)

Evaluate the integral

Unfortunately, this integral has no elementary antiderivative, suggesting either further simplification is needed or it might be a special integral that can be looked up in integral tables. In this case, the integral computed in step 4 is best evaluated numerically (approximated) or may be left in its current form if an exact answer is not required.