Section 14.5 Homework

From Calculus 10e by Larson and Edwards, p. 1007. Exercises 5, 7

Exercise 14.5.5 Finding Surface Area

Find the area for the surface given by $z=f(x,y)$ over the region $R$.

$f(x,y)=9-x^{2}$

$R$: square with vertices (0,0), (2,0), (0,2), (2,2).
Solution Since $f_{x}(x,y)=-2x$ and $f_{y}(x,y)=0$, the area is

$$S $$	$$= \int_{R} \int \sqrt{ 1 + [f_{x}(x_{i},y_{i})]^{2} + [f_{y}(x_{i},y_{i})]^{2} } \: d A$$
	$$= \int_{R} \int \sqrt{ 1+ 4x^{2}} \:d A.$$

The bounds for $R$ are $0 \leqslant x \leqslant 2$ and $ 0 \leqslant y \leqslant 2$. The integral becomes

$$ S$$	$$= \int_{0}^{2} \int_{0}^{2} \sqrt{1+4x^{2}} \:dy \: dx$$
	$$= \left. \int_{0}^{2} y\sqrt{1+4x^{2}}\right]_{0}^{2} \: dx $$
	$$= 2\int_{0}^{2} \sqrt{1+4x^{2}} \: dx$$
	let $u=2x$ and $a = 1$
	$$= 2 \left[ \frac{1}{2} \left( 2x \sqrt{1+(2x)^{2}} + \ln \left(2x + \sqrt{1+(2x)^{2}} \right) \right) \right]_{0}^{2}$$
	$$= 2 \left[ \frac{1}{2} \left( 4 \sqrt{17} + \ln \left(4 + \sqrt{17} \right) \right) \right]$$
	$$= 4 \sqrt{17} + \ln (4 + \sqrt{17})$$

Find the area for the surface given by $z=f(x,y)$ over the region $R$.

$f(x,y)= 3 +x^{3/2}$

$R$ rectangle with vertices (0,0), (0,4), (3,4), (3,0).

Solution Since $f_{x}(x,y)=\frac{3}{2}x^{1/2}$ and $f_{x}(x,y)=0$, the area is

$$S $$	$$= \int_{R} \int \sqrt{ 1 + [f_{x}(x_{i},y_{i})]^{2} + [f_{y}(x_{i},y_{i})]^{2} } \: d A$$
	$$= \int_{R} \int \sqrt{ 1+ \left( \frac{3}{2}x^{1/2} \right)^{2}} \:d A.$$

The bounds for $R$ are $0 \leqslant x \leqslant 4$ and $ 0 \leqslant y \leqslant 3$. The integral becomes

$$ S$$	$$= \int_{0}^{4} \int_{0}^{3} \sqrt{1+\frac{9}{4}x} \:dy \: dx$$
	$$= \left. \int_{0}^{4} y\sqrt{1+\frac{9}{4}x} \right]_{0}^{3} \: dx $$
	$$= 3 \int_{0}^{4} \sqrt{1+\frac{9}{4}x} \: dx $$
	Let $u=\frac{3}{2}\sqrt{x}$ and $a=1$.
	$$= 3 \left[ \frac{1}{2} \left(\frac{3}{2}\sqrt{x}\sqrt{1+\frac{9}{4}x} + \ln\left( \frac{3}{2}\sqrt{x} + \sqrt{1+\frac{9}{4}x}\right) \right) \right]_{0}^{4} $$
	$$= 3 \left[ \frac{1}{2} \left(\frac{3}{2}(2)\sqrt{1+\frac{9}{4}(4)} + \ln\left( \frac{3}{2}(2) + \sqrt{1+\frac{9}{4}(4)}\right) \right) \right] $$
	$$= 3 \left[ \frac{1}{2} \left( 3\sqrt{10} + \ln\left( 3 + \sqrt{10}\right) \right) \right] $$
	$$= \frac{3}{2} \left( 3\sqrt{10} + \ln\left( 3 + \sqrt{10}\right) \right) $$