In examinations, questions from Multivariable Calculus-II are often asked in the form of evaluating double and triple integrals, changing the order of integration, and applying transformations like polar, cylindrical, or spherical coordinates. Unlike standard single-variable calculus, Multivariable Calculus II deals with functions of several variables and their behavior over regions in two or three dimensions.
Short Questions on Multivariate Calculus - II
Question 1: Evaluate the improper integral:
Question 2: Compute the integral:
Question 3: Show that Γ(1/2) = √π
Question 4: Find the volume of the solid obtained by rotating y = x2, 0 ≤ x ≤ 1, about the x-axis.
Question 5: Determine whether the integral converges or diverges:
Question 6: Express the integral
Question 7: Find the area of the surface generated by revolving y =
Question 8: Evaluate the Beta function B(1/2,1/2).
Question 9: Determine whether the integral converges or diverges:
Question 10: Using Dirichlet’s integral, evaluate:
Long Type Question on Multivariate Calculus - II
Question 11: Let y =
Question 12: Evaluate
Check if you were right - full answer with solution below.
Solution 1:
\int_0^\infty (x+1)^2 \, dx
= \lim_{a \to \infty} \int_0^a (x^2 + 2x + 1) \, dx
= \lim_{a \to \infty} \left[ \frac{x^3}{3} + x^2 + x \right]_0^a
= \lim_{a \to \infty} \left( \frac{a^3}{3} + a^2 + a \right) = \infty Hence , the integral diverges
Solution 2:
Consider a parameter a > 0 and define:
I(a) = \int_0^\infty \frac{\sin(ax)}{x} \, dx Differentiate I(a) with respect to a
\frac{dI}{da} = \int_0^\infty \frac{\partial}{\partial a} \left( \frac{\sin(ax)}{x} \right) dx= \int_0^\infty \cos(ax) \, dx Actually, more rigorously, we use a limit with convergence factor e-bx:
I(a) = \lim_{b \to 0^+} \int_0^\infty \frac{\sin(ax)}{x} e^{-bx} \, dx Substitute x = t/a
I(a) = \lim_{b \to 0^+} \int_0^\infty \frac{\sin t}{t} e^{-b t / a} \, dt= \int_0^\infty \frac{\sin t}{t} \, dt \quad \text{(as \(b \to 0\))}
\int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2}
Solution 3:
\Gamma\left(\frac{1}{2}\right) = \int_0^\infty x^{-1/2} e^{-x} \, dx Make a subtitution: x = t2
dx = 2t dt
Make subsitution :
x−1/2 = (t2)−1/2 = t−1
\Gamma\left(\frac{1}{2}\right) = \int_0^\infty x^{-1/2} e^{-x} \, dx = \int_0^\infty t^{-1} e^{-t^2} \cdot 2t \, dt
\Gamma\left(\frac{1}{2}\right) = 2 \int_0^\infty e^{-t^2} \, dt Evaluate the Gaussian integral:
\int_0^\infty e^{-t^2} \, dt = \frac{\sqrt{\pi}}{2}
\Gamma(1/2) = 2 \cdot 2\pi = \pi
Solution 4:
Volume of revolution about the x-axis =
V = \pi \int_a^b [y(x)]^2 \, dx Here y = x2 , a= 0 , b = 1:
V =
\pi \int_0^1 (x^2)^2 dx
\pi \int_0^1 x^4 \, dx \\[1mm]
= \pi \left[ \frac{x^5}{5} \right]_0^1 \\[1mm]
= \pi \cdot \frac{1}{5} \\[1mm] V =
\frac{\pi}{5}
Solution 5:
The integrand has a singularity at x = 1 ( ln 1 = 0). Consider the integral near x = 1+:
\int_1^{1+\epsilon} x \ln x \, dx \approx \int_0^{\ln(1+\epsilon)} t \, dt This is divergent (logarithmic divergence).
The integral diverges.
Solution 6:
B(m,n) = 2 \int_0^{\pi/2} (\sin \theta)^{2m-1} (\cos \theta)^{2n-1} \, d\theta Comapring Exponets":
2m - 1 = 3
m = 22n - 1 = 2
n = 3/2
\int_0^{\pi/2} \sin^3 \theta \, \cos^2 \theta \, d\theta = \frac{1}{2} B\left(2, \frac{3}{2}\right) 2B(2,3/2)
Solution 7:
Surface Area Formula:
A = 2\pi \int_0^1 y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx y = √x
= dxdy = 1/2√x
A = 2\pi \int_0^1 x \sqrt{1 + (2x)^2} \, dx = 2\pi \int_0^1 x \sqrt{4x^2 + 1} \, dx
A = \pi \int_0^1 (4x + 1) \, dx Substitute t = 4x+1
dx = dt/4 :
A = 4\pi \int_1^5 t \, dt = 4\pi \cdot \frac{3}{2} \left(5^{3/2} - 1\right)
Solution 8:
B(p,q) =
\frac{\Gamma(p)\,\Gamma(q)}{\Gamma(p+q)}
B\left(\frac{1}{2}, \frac{1}{2}\right)
= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}
= \frac{\sqrt{\pi} \cdot \sqrt{\pi}}{1} =
\pi
Solution 9:
Use integration by parts. Let u = x, dv = e−xdx. Then du = dx , v = −e−x.
\int_{0}^{\infty} x e^{-x}\,dx = \left[-x e^{-x}\right]_{0}^{\infty} + \int_{0}^{\infty} e^{-x}\,dx The boundary term limx→ ∞xe−x = 0 and at 0 it is 0. Thus
= 0 +
\left[-e^{-x}\right]_{0}^{\infty} = 1So the integral converges and its value is 1. (Equivalently this is Γ(2)=1!)
Solution 10:
Let I(α) =
\int_0^\infty x e^{-\alpha x}\sin(5x)\,dx I α>0.Since J(α) =
\int_0^\infty e^{-\alpha x}\sin(5x)\,dx =\frac{5}{\alpha^2+25} , we haveI(α) =
-\frac{dJ(\alpha)}{d\alpha} =\frac{10\alpha}{(\alpha^2 + 25)^2}
J'(\alpha) = -\frac{dJ(\alpha)}{d\alpha} = \frac{10\alpha}{(\alpha^2 + 25)^2} Take α → 0+ (Abel limit):
\int_{0}^{\infty} x \sin(5x)\,dx =\lim_{\alpha \to 0^{+}} I(\alpha) \Iota = 0
Solution 11:
Using the disk method:
V = \pi \int_{0}^{4} y^{2}\,dx
= \pi \int_{0}^{4} x\,dx
= \pi \left[ \frac{x^{2}}{2} \right]_{0}^{4}
= \pi \cdot \frac{16}{2}
= 8\pi Surface area:
S = 2\pi \int_{0}^{4} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dx Here y =
\sqrt{x} \dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}} . Then
1 + \left(\frac{dy}{dx}\right)^2 =1 + \left(\frac{dy}{dx}\right)^2
= \frac{4x + 1}{1} So,
y \sqrt{1 + (y')^2} = \sqrt{x} \, \sqrt{4x + 1} = \sqrt{x(4x + 1)} Hence,
S = 2\pi \int_{0}^{4} \sqrt{4x+1}\, dx = \pi \int_{0}^{4} (4x+1)\, dx Let u = 4x+1.u = 4x+1 du = 4 dx, x = 0 ↦ u = 1 x = 4 ↦ u = 17. Then
S = \pi \int_{1}^{17} u^{1/2} \, 4\,du = 4\pi \int_{1}^{17} u^{1/2}\, du
= 4\pi \left[ \frac{2}{3} u^{3/2} \right]_{1}^{17} = \frac{8\pi}{3} \left( 17^{3/2} - 1 \right)
Solution 12:
\int_0^\infty \frac{x \sin x}{x^2 + a^2} \, dx
F(b) = \int_0^\infty \frac{\cos(bx)}{x^2 + a^2} \, dx
F(b) = \frac{\pi}{2a} e^{-ab} a > 0Differentitae w.r.t b
F'(b)= \frac{d}{db} \int_0^\infty \frac{\cos(bx)}{x^2 + a^2} \, dx =
\int_0^\infty \frac{\partial}{\partial b} \frac{\cos(bx)}{x^2 + a^2} \, dx =
\int_0^\infty \frac{x \sin(bx)}{x^2 + a^2} \, dx Differentiate the closed form:
F'(b) = \frac{d}{db} \left( \frac{\pi}{2a} e^{-ab} \right) = - \frac{\pi}{2} e^{-ab} =
\int_0^\infty \frac{x \sin(bx)}{x^2 + a^2} \, dx = \frac{\pi}{2} e^{-ab} For b = 1
\int_0^\infty \frac{x \sin x}{x^2 + a^2} \, dx = \frac{\pi}{2} e^{-a}