Solve ∫ (from 0 to 2 pi) of int_0^7(sinr^2)rdrdθ

$$\int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 7 } ( \sin r ^ { 2 } ) r d r d \theta$$

Evaluate

$$\pi \left(1-\cos(49)\right)\approx 2.197253326$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \int _{0}^{7}\sin(r^{2})r\mathrm{d}r\mathrm{d}\theta$$

Find the integral of \int _{0}^{7}\sin(r^{2})r\mathrm{d}r using the table of common integrals rule \int a\mathrm{d}\theta =a\theta .

$$\int _{0}^{7}\sin(r^{2})r\mathrm{d}r\theta$$

Simplify.

$$\frac{1}{2}\left(2\left(-\frac{\cos(49)}{2}\right)+1\right)\theta$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$2\left(-\frac{1}{2}\cos(49)+\frac{1}{2}\right)\pi +0\left(-\frac{1}{2}\cos(49)+\frac{1}{2}\right)$$

Simplify.

$$\pi \left(-\cos(49)+1\right)$$