\(\int_0^2a \int_0^{x^2/4a} x.y dxdy=\int_0^{2a} xdx \int_0^{x^2/4a} ydy=\int_0^{2a}\frac{xx^4}{32a^2}dx\) =\(\frac{1}{32a^2} \int_0^{2a}x^5 dx=\frac{a^4}{3}\)

Given, F(x)=\(\int_0^1\int_{x^2}^x xy(x+y)dydx=\int_0^1 \int_{x^2}^x(x ^2 y+xy^2)dydx\)
=\(\int_0^1 [\frac{x^2 y^2}{2}+\frac{xy^3}{3}] _x^{x^2}dx=\int_0^1 [\frac{x^3}{2}+\frac{x^4}{3}-\frac{x^4}{2}-\frac{x^5}{3} ]dx=\frac{1}{2}+\frac{1}{3}-\frac{1}{2}-\frac{1}{5}=\frac{2}{15}\)

Given, f(x)=\(\int_0^1 \int_0^{1-y} xy\sqrt{1-x-y} dxdy\)
putting,t=x/(1-y)=>x=t(1-y)=>dx=(1-y)dt
\(\int_0^1 \int_0^1 t(1-y)y\sqrt{1-t(1-y)-y} (1-y)dtdy\)
=\(\int_0^1 \int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy\)
=\(\int_0^1y(1-y)^{5/2} dy \int_0^1 t(1-t)^{1/2} dt\)
=\(\int_0^1 y^{2-1} (1-y)^{7/2-1} dy\int_0^1 t^{2-1} (1-t)^{3/2-1} dt=\beta(2,\frac{7}{2})\beta(2,\frac{3}{2})=\frac{16}{945}\)

Consider case of r=R
Then,
D=D?
Where D? is surface density of planet
As D → ∞, r → 0
For finding mass of planet, we use triple integration formula
M=∫∫∫ dV
Converting into spherical co-ordinates, we get
M=\(\int\int\int D’r^2 sin\ta \frac{R+r}{2r} dr d\ta d\ta\)
Applying limits
0 to π
0 to 2π
0 to R
Solving Triple Integral we get,
M=\(\frac{5\pi D’R^3}{3}\)
Thus, mass of planet is \(\frac{5\pi D’R^3}{3}\).

The above statement is not necessarily true always but change of variables cartesian to polar form & changing into suitable form reduces complexity in evaluation of double integral however converse is also true for example ?_R x² y³ dx dy is given integral while solving re is no need of converting it into polar form since ?_P r⁶ cos² ? sin³ ? dr d? is tedious to solve when R only consists of constants but ?_R x² y³ dx dy can be solved easily by ordinary method of integration.

Given, f(x)=\(\int_1^0?_0^{1-y} xy\sqrt{1-x-y} \,dxdy\)
putting, \(t=\frac{x}{1-y}\)=>x=t(1-y)=>dx=(1-y)dt
\(\int_1^0\int_0^1 t(1-y)y\sqrt{1-t(1-y)-y} (1-y)dtdy\)
=\(\int_1^0\int_0^1 y(1-y)^{5/2} t(1-t)^{1/2} dtdy\)
=\(\int_1^0y(1-y)^{5/2} dy \int_0^1 t(1-t)^{1/2} dt\)
=\(\int_0^1 y^{2-1} (1-y)^{7/2-1} dy \int_0^1 t^{2-1} (1-t)^{3/2-1} dt=?(2,\frac{7}{2})?(2,\frac{3}{2})=\frac{16}{945}\)

Add constant automatically
Given, f(x)=\(\int_0^1 \int_0^x (x^2+y^2) dydx = \int_0^1 (x^3+\frac{x^3}{3})dx=1+\frac{1}{3}=\frac{2}{3}\)

Add constant automatically
∫x³ Sin(x)dx = -x³ Cos(x) + 3∫x² Cos(x)dx
∫x² Cos(x)dx = x² Sin(x) – 2∫xSin(x)dx
∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x)
=> ∫x³ Sin(x)dx = -x³ Cos(x) + 3[x² Sin(x) – 2[-xCos(x) + Sin(x)]] => ∫x³ Sin(x)dx = -x³ Cos(x) + 3x² Sin(x) + 6xCos(x) – 6Sin(x)
and, ∫y³ Sin(y)dy = -y³ Cos(x) + 3∫y² Cos(y)dy
∫y² Cos(y)dy = y² Sin(y) – 2∫ySin(y)dy
∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y)
=> ∫y³ Sin(y)dy = -y³ Cos(y) + 3[y² Sin(y) – 2[-yCos(y) + Sin(y)]] => ∫y³ Sin(y)dy = -y³ Cos(y) + 3y² Sin(y) + 6yCos(y) – 6Sin(y)
Hence, ∫∫x³ y³ sin?(x) sin?(y) dxdy = (∫x³ Sin(x)dx)(∫y³ Sin(y)dy) = (-x³ Cos(x) + 3x² Sin(x)+6xCos(x) – 6Sin(x))(-y³ Cos(y) + 3y² Sin(y) + 6yCos(y) – 6Sin(y)).

We know that volume of sphere is,
\(V = \frac{4}{3} ?R^3 \)
Differentiating above equation with respect to R we get,
\(\frac{dV}{dR}= \frac{4}{3} ?�3R^2=4?R^2\)
Since volume of sphere should not exceed more than 4%,
\(dR=\frac{dV}{4?R^2}=\frac{0.04}{4?(5)^2}=0.000127\)
Error in radius = 0.0127%

\(\int_{x=0}^1 \int_{y=0}^{1-x} \int_{z=0}^{1-x-y} \frac{dz dy dx}{(1+x+y+z)^3} = \int_0^1 \int_0^{1-x}\Big[\frac{-1}{(2(1+x+y+z)^2}\Big]_{z=0}^{1-x-y} \,dy \,dx\)
\(\int_0^1 \int_0^{1-x}[\frac{-1}{8} + \frac{1}{(2(1+x+y)^2)}] \,dy \,dx = \int_0^1 \Big[\frac{-y}{8} – \frac{1}{2(1+x+y)}\Big]_{y=0}^{1-x} \,dx\)
\(\int_{x=0}^1 \Big[\frac{-(1-x)}{8} – \frac{1}{4} + \frac{1}{2(x+1)}\Big]dx = \int_{x=0}^1 \Big[\frac{-3}{8} + \frac{x}{8} + \frac{1}{2(x+1)}\Big]dx\)
\(\Big[\frac{-3x}{8} + \frac{x^2}{16} + \frac{1}{2} log(x+1)\Big]_{x=0}^1 = \frac{3}{8} + \frac{1}{16} + \frac{log?2}{2} = log\sqrt{2} – \frac{5}{16}.\)