Q16Mathematics I

Question

Evaluate the integral $\int_{0}^{1} \int_{0}^{x} \frac{x ^{3} d y d x}{x ^{2} + y ^{2}}$ by changing into polar coordinates.

Answer 1

The mathematical problem demands the rigorous evaluation of the specific definite double integral:

I = \int_{0}^{1} \int_{0}^{x} \frac{x ^{3}}{x ^{2} + y ^{2}} d y d x

Direct integration in Cartesian coordinates $(x, y)$ is mathematically cumbersome due to the radical in the denominator. Converting the entire system to polar coordinates $(r, θ)$ drastically simplifies the geometric integrand.

Step 1: Coordinate Transformation

The fundamental polar substitutions are $x = r cos θ$ and $y = r sin θ$ . The denominator cleanly simplifies: $x^{2} + y^{2} = r^{2} = r$ . The differential area element transforms rigorously to $d y d x = r d r d θ$ .

We must carefully map the original 2D region. The inner limit $y$ goes from $0$ to $x$ , describing a right triangle bounded by the x-axis and the $y = x$ line (which is exactly $θ = π /4$ ). The outer limit $x$ goes from $0$ to $1$ . The vertical line $x = 1$ in polar terms is $r cos θ = 1$ , or $r = sec θ$ .

The fully transformed polar integral is:

I = \int_{0}^{π /4} \int_{0}^{s e c θ} \frac{( r cos θ ) ^{3}}{r} (r d r d θ)

We algebraically cancel the $r$ terms:

I = \int_{0}^{π /4} \int_{0}^{s e c θ} r^{3} cos^{3} θ d r d θ

Step 2: Execution of Integration

We integrate first strictly with respect to $r$ :

I = \int_{0}^{π /4} cos^{3} θ [\frac{r ^{4}}{4}]_{0}^{s e c θ} d θ = \frac{1}{4} \int_{0}^{π /4} cos^{3} θ sec^{4} θ d θ

Because $sec θ = 1/ cos θ$ , the trigonometric term $cos^{3} θ sec^{4} θ$ mathematically reduces entirely to $sec θ$ :

I = \frac{1}{4} \int_{0}^{π /4} sec θ d θ

The standard antiderivative of $sec θ$ is known rigorously to be $ln ∣ sec θ + tan θ ∣$ :

I = \frac{1}{4} [ln ∣ sec θ + tan θ ∣]_{0}^{π /4}

We evaluate at the upper bound $π /4$ , where $sec (π /4) = 2$ and $tan (π /4) = 1$ :

Upper = ln (2 + 1)

We evaluate at the lower bound $0$ , where $sec (0) = 1$ and $tan (0) = 0$ :

Lower = ln (1 + 0) = ln (1) = 0

Thus, the final, exact mathematically verified result is unequivocally $\frac{1}{4} ln (2 + 1)$ .

Q16Mathematics I

Question

Answer

Step 1: Coordinate Transformation

Step 2: Execution of Integration