$f(x,y)=4y+3x^2$ between $y=2x,y=0$ and $x=2$.

Question

To find the surface area of the portion of the surface $f(x,y)=4y+3x^2$ between $y=2x,y=0$ and $x=2$.

The limits of $y$ is clear. What is the limit of $x$? Is it from $x=0$ to $x=2$

Answer 1

Yes. The lines $y=0$ and $y=2x$ meet at $(0,0)$. So, the region bounded by the previous two lines and by the line $x=2$ is the triangle whose vertices are $(0,0)$, $(2,0)$, and $(2,4)$. So, your integral is$$\int_0^2\int_0^{2x}4y+3x^2\,\mathrm dy\,\mathrm dx.$$

Answer 2

The domain of integration is $$\{(x,y)\in \Bbb R^2 \;\; : \; 0\le x\le 2 \text{ and } 0\le y\le 2x\}$$

the integral becomes

$$\int_0^2\int_0^{2x}(4y+3x^2)dxdy=$$

$$\int_0^2\Bigl(\int_0^{2x}(4y+3x^2)dy\Bigr)dx=$$ $$\int_0^2\Bigl[2y^2+3x^2y\Bigr]_0^{2x}dx=$$ $$\frac{136}{3}$$

Answer 3

Presumably you want the area of the surface $z=f(x,y)$, also known as the graph of $f(x,y)$. Anyway, you got the correct limits. The region is a triangle defined by the inequalities $0\le y\le 2x, 0\le x\le2$.

The surface area $A$ of the graph of $f(x,y)$ with $(x,y)$ ranging over a region $S$ is calculated by the integral $$ A=\int_S\sqrt{1+f_x^2+f_y^2}\,dy\,dx. $$ Here $f_x(x,y)=6x$ and $f_y(x,y)=4$, so the integrand is $\sqrt{17+36x^2}$. This doesn't depend on $y$ at all, which indicates that it may be simpler to integrate over $y$ first. So $$ A=\int_{x=0}^2\int_{y=0}^{2x}\sqrt{17+36x^2}\,dy\,dx. $$ The inner integral is $$ \int_{y=0}^{2x}\sqrt{17+36x^2}\,dy=2x\sqrt{17+36x^2}. $$ Leaving the outer integral to you. You are expected to spot that the outer integrand is (up to a scalar multiple) of the form $g'(x)g(x)^\alpha$ for a suitable function $g(x)$ and exponent $\alpha$.