Question

Let C be the curve x = 4 - $y^2$ from (0, -2) to (0,2). Then the value of integral ${\int }_C(y^{3}dx+x^{2}dy)$ will be

• A. $\frac{64}{15}$
• B. 0
• C. $\frac{128}{15}$
• D. $\frac{72}{5}$

Answer 1

The correct option is C $\frac{128}{15}$

Given, equation of curve $x=4-y^2$


$I={\int }_C(y^{3}dx+x^{2}dy)$

$\because$$x=4-y^2$

$\therefore$ dx = -2y dy

So, $I={\int }_{-2}^2\left[y^3\right(-2ydy)+(4-y^2{)}^{2}dy]$

$={\int }_{-2}^2[-2y^4+16+y^4-8y^2]dy$

$={\int }_{-2}^2(16-8y^2-y^4)dy$

$=2{\int }_0^2(16-8y^2-y^4)dy$

$=2{\left[16\right(y)-\frac{8y^3}{3}-\frac{y5}{5}]}_0^2$

$=2\left[16\right(2)-\frac{8}{3}(8)-\frac{1}{5}(32\left)\right]$

$=2[32-\frac{64}{3}-\frac{32}{5}]=\frac{128}{15}$

Alternate Method:

Given curve 'C' formed a closed region from (0, -2) to (0, 2)

$I={\oint }_C{y}^{3}dx+x^{2}dy$

By Green theorem,

${\oint }_{C}Mdx+Ndy=\int {\int }_R(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dxdy$

$=\int {\int }_R(2x-3y^2)dxdy$

$={\int }_{-2}^2{\int }_{x=0}^{x=4-y^2}(2x-3y^2)dxdy$

$={\int }_{-2}^2{\int }_{x=0}^{x=4-y^2}(x^2-3y^{2}x)dy$

$={\int }_{-2}^2\left[\right(4-y^2{)}^2-3y^2(4-y^2)]dy$

$={\int }_{-2}^2(16+y^4-8y^2-12y^2+3y^4)dy$

$={\int }_{-2}^2(4y^4-20y^2+16)dy$

$=2{[\frac{4y^5}{5}-\frac{20y^3}{3}+16y]}_0^2$

$=2[\frac{4}{5}\times 32-\frac{20}{3}\times 8+32]=\frac{128}{15}$