Question

Each integral represents the volume of a solid. Describe the solid.

(a) \(\pi \int_0^{\pi /2} {{{\cos }^2}} xdx\)

Short answer

When this region is rotated about \(x\)-axis, a solid hemisphere of radius 1 with a conical tip is obtained.

Step-by-step solution

Volume of a solid

Consider a solid that lies between the curves\(x = a\), and\(x = b\). If the cross-section of\(S\)in the plane\({P_x}\), through\(x\)and perpendicular to the\(x\)-axis, is given by an integrable function\(A(x)\), then the volume of\(S\)is\(\begin{aligned}{}V = \mathop {\lim }\limits_{\max \Delta {x_i} \to 0} \sum\limits_{i = 1}^n A \left( {{x_i}^*} \right)\Delta {x_i}\\V = \int_a^b A (x)dx.\end{aligned}\)

Compare the given equation \(\pi \int_0^{\pi /2} {{{\cos }^2}} xdx\) with the formula of volume

It is given that the integral is \(\pi \int_0^{\frac{\pi }{2}} {{{\cos }^2}} xdx\).

On comparing with the formula, \(A(x) = \pi {\cos ^2}x\) which is further written as \(A(x) = \pi {\cos ^2}x = \pi {y^2}\).

Thus, the curve becomes \(y = \cos x\).

As the value of \(x\) varies from 0 to \(\frac{\pi }{2}\), then the value of \(y\) varies from 1 to 0.

The region bounded under the curve is a quarter with one side of 1 and other side of \(\frac{\pi }{2}\).

When this region is rotated about \(x\)-axis, a solid hemisphere of radius 1 with a conical tip is obtained.